A trigonometric equation of the form A cos θ + B sin θ = k does not always have a real solution. The expression on the left has amplitude R = √(A² + B²), so its values stay between −R and R. When θ is allowed to cover a full period, the equation has a solution exactly when |k| ≤ R.
For example, 3cos θ + 4sin θ = k has amplitude 5. Therefore real solutions exist over a full period precisely when −5 ≤ k ≤ 5. For −5 < k < 5 there are two solutions in a standard half-open full-period interval; at k = 5 or −5 there is one. This guide shows how to derive the condition, count solutions, handle parameters inside the coefficients and avoid applying a full-period range to a restricted angle interval.
Why the equation may have no solution
Sine and cosine values are bounded: each lies between −1 and 1. Their weighted sum is also bounded, though its maximum is not generally |A| + |B|.
For instance, 3cos θ + 4sin θ cannot reach an output of 10, regardless of θ. The exact maximum is 5, as will follow from a single-wave representation.
Therefore an equation A cos θ + B sin θ = k is partly a range question before it is a trigonometric-solving question.
If the right-hand value lies outside the possible range, attempting inverse sine or inverse cosine will either fail or produce an invalid argument. Checking existence first saves unnecessary calculation and prevents an invented angle from becoming the final answer.
The amplitude formula R = √(A² + B²)
For fixed real A and B, not both zero, the expression A cos θ + B sin θ can be written as R cos(θ − α) for a suitable phase α.
Expanding gives R cosθ cosα + R sinθ sinα. To match the original coefficients, choose R cosα = A and R sinα = B.
Square and add these equations. Since cos²α + sin²α = 1, we obtain R² = A² + B², so R = √(A² + B²) > 0.
Because cosine ranges from −1 to 1 over a full cycle, the weighted sum ranges exactly from −R to R. This is not merely an upper estimate; both extremes are attained for suitable angles when the full period is available.
Work the 3–4–5 example
For 3cos θ + 4sin θ, calculate R = √(3² + 4²) = √(9 + 16) = 5.
Choose α such that cosα = 3/5 and sinα = 4/5. Then 3cos θ + 4sin θ = 5cos(θ − α).
Thus the original equation becomes 5cos(θ − α) = k, or cos(θ − α) = k/5.
Since a real cosine value must lie between −1 and 1, require −1 ≤ k/5 ≤ 1. Multiplying by the positive 5 gives −5 ≤ k ≤ 5.
This is the complete existence condition when θ may range across a full period. A longer trigonometric solution is needed only when the question asks for the actual angles.
Why −5 and 5 are included
At k = 5, the equation becomes cos(θ − α) = 1. This is satisfied when θ = α modulo 2π.
At k = −5, the equation becomes cos(θ − α) = −1. This holds when θ = α + π modulo 2π.
Both values are actually attained by the weighted sine–cosine expression. Therefore the correct parameter range is inclusive: −5 ≤ k ≤ 5.
A student who writes −5 < k < 5 has incorrectly excluded the maximum and minimum values. Those boundary cases produce fewer distinct angles per period, but they are not impossible.
Count solutions in a half-open full period
Use the standard interval 0 ≤ θ < 2π, which represents one full cycle without counting the same direction twice at the endpoints.
For −5 < k < 5, the equation cos(θ − α) = k/5 has exactly two distinct solutions in one full cycle.
For k = 5, there is one solution modulo 2π corresponding to the cosine maximum. For k = −5, there is one corresponding to the cosine minimum.
For |k| > 5, there are no real solutions. This count is tied to a complete period and the given half-open interval; a shorter or differently bounded interval may contain fewer solutions.
What happens when k = 0?
Consider 3cos θ + 4sin θ = 0. In R-form, this is cos(θ − α) = 0.
Cosine vanishes when its argument equals π/2 or 3π/2 modulo 2π. Hence in one complete period, the angles are θ = α + π/2 and θ = α + 3π/2, interpreted modulo 2π.
With α = arctan(4/3), approximately 53.130°, these are approximately 143.130° and 323.130° in 0° ≤ θ < 360°.
The equation has two solutions, consistent with the strict interior condition −5 < 0 < 5. The exact phase representation is more informative than trusting a rounded inverse-tangent value alone.
A parameter within the sine coefficient
Suppose the equation is 3cos θ + m sin θ = 5, where m is real. We want to know for which m at least one real θ exists over a full period.
Its amplitude is R = √(3² + m²) = √(9 + m²). Existence requires |5| ≤ R, so 5 ≤ √(9 + m²).
Both sides are non-negative, so squaring is safe: 25 ≤ 9 + m². Therefore m² ≥ 16.
Hence m ≤ −4 or m ≥ 4. This condition is necessary and sufficient when the angle may cover a full period, because the expression attains every value between minus and plus its amplitude.
What occurs at the parameter boundaries
For m = 4, the equation becomes 3cos θ + 4sin θ = 5. Its amplitude is 5, so the right-hand value is exactly the maximum.
There is one distinct solution in 0 ≤ θ < 2π, occurring at the phase angle where the weighted sum reaches that maximum.
For m = −4, the equation becomes 3cos θ − 4sin θ = 5. Its amplitude is also 5, and again there is one maximum point per full cycle.
For |m| > 4, the amplitude exceeds 5, so the right-hand side lies strictly between the maximum and minimum. There are then two solutions per full period. For |m| < 4, the amplitude is too small and no solution exists.
The sign of m changes phase, not amplitude
The two equations 3cos θ + 4sin θ = k and 3cos θ − 4sin θ = k have the same amplitude √(9 + 16) = 5.
Their peaks occur at different angles because changing the sign of the sine coefficient changes the phase. But their full-period ranges are both [−5, 5].
Therefore the existence condition for a fixed right-hand k is unchanged when B is replaced by −B. The number of full-period solutions at interior and boundary levels is also unchanged.
This is useful when a parameter may be positive or negative. The amplitude depends on its square, while the angle at which an extreme occurs depends on its signed value.
A second parameter example involving both sides
Consider m sin θ + 4cos θ = m + 1. Find real m for which some θ exists over a full period.
The amplitude of the left side is √(m² + 16). The right side is m + 1, which may be positive or negative. The existence condition is |m + 1| ≤ √(m² + 16).
Both sides are non-negative, so square safely: (m + 1)² ≤ m² + 16. Expand to m² + 2m + 1 ≤ m² + 16.
Cancel m² to obtain 2m ≤ 15, hence m ≤ 15/2. Every real m at or below 7.5 satisfies the full-period existence condition; m above 7.5 does not.
Check an inside and an outside parameter
Take m = 0 in the previous equation. It reduces to 4cos θ = 1. Since 1 lies between −4 and 4, solutions exist. The condition m ≤ 15/2 correctly includes zero.
Take m = 10. The equation becomes 10sin θ + 4cos θ = 11. Its amplitude is √116, approximately 10.770, less than 11, so no real solution exists.
At m = 15/2, the right-hand side equals the amplitude. This is a boundary maximum case with exactly one solution per full period.
These checks support the derived range without replacing the algebraic proof that |m + 1| ≤ √(m² + 16) is equivalent to m ≤ 15/2.
Why the absolute value around k matters
An equation R cos(θ − α) = k is solvable over a full period when −R ≤ k ≤ R. This is equivalent to |k| ≤ R.
If k is an expression involving a parameter, it may be negative. Replacing |k| with k without justification can produce a weaker condition that permits impossible large negative outputs.
For example, m sin θ + 4cos θ = m + 1 has a right-hand side that can be negative when m < −1. The correct existence condition still compares its magnitude with the amplitude.
Squaring both sides of |m + 1| ≤ √(m² + 16) is legitimate because both sides are non-negative. The absolute-value formulation makes that logic explicit.
A bound from separate sine and cosine limits is too weak
Because |sin θ| ≤ 1 and |cos θ| ≤ 1, one might write |3cos θ + 4sin θ| ≤ 3 + 4 = 7.
That bound is true but not sharp. It would incorrectly suggest that k = 6 might be attainable, even though the exact amplitude is 5.
The problem is that sine and cosine cannot both equal 1 at the same angle. Their values are linked by sin²θ + cos²θ = 1.
The R-form calculation respects that relationship and gives the exact maximum √(3² + 4²) = 5. Exact amplitude, not a loose sum of magnitudes, is what determines whether a particular equation is solvable.
A geometric proof using the unit circle
The point (cos θ, sin θ) lies on the unit circle. The expression A cos θ + B sin θ is the dot product of the vector (A, B) with that unit vector.
By the Cauchy–Schwarz inequality, its magnitude cannot exceed √(A² + B²) × 1 = √(A² + B²).
Equality occurs when the unit vector points parallel or opposite to (A, B), so the maximum and minimum are actually reached over a full rotation.
This gives a geometric explanation of the R-form result. The unit circle restricts which pairs of sine and cosine values can occur simultaneously, and that restriction produces the Euclidean amplitude rather than |A| + |B|.
How to find actual θ-values after checking existence
Once |k| ≤ R has been established, write A cos θ + B sin θ = R cos(θ − α), with cosα = A/R and sinα = B/R.
Then solve cos(θ − α) = k/R. Let β = arccos(k/R), with 0 ≤ β ≤ π. The general solutions are θ − α = ±β + 2πn, with integer n.
Therefore θ = α ± β + 2πn. Filter these values into the interval stated by the question.
This two-stage method separates the existence decision from listing angles. If a problem asks only for the allowed parameter values, there is often no need to calculate α or β numerically.
A simple exact-angle case
For 3cos θ + 4sin θ = 3, let α be the first-quadrant angle satisfying cosα = 3/5 and sinα = 4/5.
The equation becomes 5cos(θ − α) = 3, hence cos(θ − α) = 3/5 = cosα.
Thus θ − α = ±α modulo 2π. One family gives θ = 0 modulo 2π; the other gives θ = 2α modulo 2π.
On 0 ≤ θ < 2π, the two solutions are θ = 0 and θ = 2arctan(4/3), approximately 106.260°. This is a useful example because one solution lies exactly at a boundary of the standard interval.
Interval endpoints can change the solution count
For the equation 3cos θ + 4sin θ = 3, the values θ = 0 and θ = 2α lie in the half-open interval 0 ≤ θ < 2π.
If the question instead specifies 0 < θ < 2π, the solution θ = 0 is excluded. The coterminal value θ = 2π is also excluded, so only θ = 2α remains.
If the interval is closed 0 ≤ θ ≤ 2π, both 0 and 2π are valid numerical angle values, together with 2α. They describe the same circle direction at the two endpoints but are distinct listed values in the stated interval.
Therefore the rule “two solutions for an interior k” must be understood as a statement about one complete half-open period. Endpoint wording can change the count.
A restricted interval has a smaller possible range
The full-period bound |k| ≤ 5 is necessary for 3cos θ + 4sin θ = k, but may not be sufficient when θ is restricted to a short interval.
Suppose 0 ≤ θ ≤ π/4. The expression increases throughout this interval because its derivative is −3sin θ + 4cos θ, which remains positive there.
At θ = 0 its value is 3. At θ = π/4 its value is 3/√2 + 4/√2 = 7/√2.
Thus its range on this restricted interval is [3, 7/√2], not [−5, 5]. For example, k = 5 satisfies the global amplitude bound but cannot occur between 0° and 45°, because the maximum there is about 4.950.
Why a global amplitude condition may be insufficient
The amplitude R describes what the wave can attain over a complete cycle. It does not guarantee that a particular maximum or minimum occurs inside a shorter permitted interval.
For 3cos θ + 4sin θ, the maximum 5 occurs at θ = α, about 53.130°, outside the interval 0° ≤ θ ≤ 45°.
Therefore the equation with k = 5 has a solution over a full rotation but none in that restricted interval. The difference arises from allowed angles, not from a change in the function.
A student should use the global amplitude bound as a quick impossibility check and then inspect the actual restricted interval when existence or solution count is asked for on less than a full period.
Double-angle arguments change how many solutions fit
Consider 3cos(2θ) + 4sin(2θ) = k for 0 ≤ θ < 2π. Its amplitude is still 5, because replacing θ with 2θ does not change the range of the weighted wave.
However, the inner angle 2θ covers two full cycles as θ runs through one standard period. Therefore for −5 < k < 5 there are four distinct θ-solutions, rather than two.
At k = ±5 there are two solutions in this interval, one maximum or minimum per inner cycle. For |k| > 5 there are none.
This distinction separates existence conditions from solution counts. The amplitude bound remains unchanged, while the number of complete oscillations inside the stated θ-interval changes.
An exact double-angle check
For 3cos(2θ) + 4sin(2θ) = 3, let α again satisfy cosα = 3/5 and sinα = 4/5.
The R-form equation becomes cos(2θ − α) = cosα. Therefore 2θ − α = ±α + 2πn.
One family gives 2θ = 2πn, so θ = πn. The other gives 2θ = 2α + 2πn, so θ = α + πn.
In 0 ≤ θ < 2π, the four distinct answers are 0, α, π and π + α. This verifies the predicted four-solution count for an interior k-value when the argument is doubled.
What if A and B are both zero?
If A = 0 and B = 0, then A cos θ + B sin θ is identically zero for every real θ. Its amplitude R is zero.
The equation 0 = k has every allowed θ as a solution when k = 0 and no solution when k ≠ 0.
The ordinary phase representation using cosα = A/R would involve division by zero, so it must not be used in this degenerate case.
This matters when A and B themselves depend on a parameter and can simultaneously vanish. Check R = 0 separately before forming a phase angle or assuming the usual one-or-two solution count.
The case of a single trigonometric term
If B = 0 but A ≠ 0, the equation is A cos θ = k. Its amplitude is |A|, so full-period existence requires |k| ≤ |A|.
If A = 0 but B ≠ 0, the equation is B sin θ = k, with the same style of condition |k| ≤ |B|.
These are direct special cases of R = √(A² + B²), so no separate unfamiliar formula is needed.
For example, 4sin θ = 5 has no real solution, while 4sin θ = 4 has one distinct solution in 0 ≤ θ < 2π. The amplitude comparison makes both decisions immediate.
A rapid checklist for parameter problems
First identify the coefficients A and B of cos θ and sin θ. Determine whether either depends on the unknown parameter.
Second compute R = √(A² + B²). If R = 0, handle the identically zero case separately.
Third impose |k| ≤ R for existence over a full period. Solve the resulting inequality carefully, checking that any squaring step has non-negative sides.
Fourth distinguish strict interior parameter values from equality cases when a solution count is requested. Finally inspect the actual angle interval; if it is shorter than a full period, the global range may be only a necessary condition.
This routine gives a complete existence answer without unnecessarily solving every trigonometric angle.
Practice questions with checked answers
Question one: 3cos θ + 4sin θ = k, with θ over a full period. Parameter condition: −5 ≤ k ≤ 5.
Question two: the same equation on 0 ≤ θ < 2π. Number of solutions: two if −5 < k < 5, one if k = ±5, none otherwise.
Question three: 3cos θ + m sin θ = 5 over a full period. Condition: m ≤ −4 or m ≥ 4.
Question four: m sin θ + 4cos θ = m + 1 over a full period. Condition: m ≤ 15/2.
Question five: sin θ + 2cos θ = 3 over a full period. Answer: no solution because √5 < 3.
Question six: 3cos(2θ) + 4sin(2θ) = k on 0 ≤ θ < 2π. Solution count: four when |k| < 5, two when |k| = 5, none when |k| > 5.
For each, state whether the question is about existence, counting or actual angles before deciding how far to work.
Common errors and the first repair
A student may use |A| + |B| as the maximum rather than √(A² + B²). Another may square a parameter inequality without first writing an absolute value around the right-hand expression.
A third may calculate the correct amplitude but claim two solutions even at the maximum, where there is only one per period. A fourth may use a global range although the permitted angle interval is too short to reach its extremes.
These are different misconceptions. A useful tutorial diagnoses whether the difficulty lies in R-form, range, strictness, periodicity or interpreting interval endpoints.
A changed equation with one coefficient negative, or with a doubled angle argument, provides a stronger transfer check than repeating the same 3–4–5 numbers.
Syllabus suitability and further reading
Trigonometric amplitude, compound-angle identities, graph ranges and parameter inequalities are interconnected Additional Mathematics skills. Some solution-count and parameter extensions may be enrichment depending on the learner’s specific syllabus and school progression.
For Singapore, consult the applicable SEAB listings for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2.
Related guides include How to Recover a Sine Graph Formula from Amplitude, Period and Midline, How to List Every Trigonometric Solution on an Interval Crossing Zero, and How to Distinguish Horizontal and Vertical Graph Shifts. Explore the Additional Mathematics archive or enquire about eduKate Singapore maximum three-student tutorials near Sixth Avenue MRT.
The lasting skill is to determine whether an equation can have a solution before solving for angles, and then distinguish the global wave range from the actual interval specified in the question. Properly taught kids shine a bright light into the future.