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Solving equations with the R formula

Solve a sin x + b cos x = c using the R formula. Transform the angle interval, find every branch, handle multiple angles and reject excluded or squared candidates.

Before you startThe R formula, inverse sine/cosine and interval root sets

01 / Reduce the equation to one trig function

A phase shift changes the interval you must search.

First combine the sine and cosine into one wave, then solve the shifted equation. The amplitude checks whether solutions are possible at all; the original interval determines which solutions belong in the final answer.

a sin(mx) + b cos(mx) = c
R sin(mx + α) = c

When R > 0, divide by R. Keep the original x interval and transform both endpoints into the new angle u = mx + α. The model shows why checking just the principal inverse sine misses roots.

Intersections in the chosen intervalExplore

Target cInput multiplier mInterval

3sin x + 4cos x = 3.

Solutions (2): 90.000°, 343.740°.

Put u = x + α, where α ≈ 53.1301°. The transformed interval is 53.1301° ≤ u < 413.1301°.

Gold points mark all solutions inside the selected interval. Angles shown in decimals are rounded to three places. The calculations use both periodic branches before filtering the interval.

02 / Check amplitude before using an inverse function

The target must lie within the wave’s global range.

For R > 0, an equation R sin u = c has real solutions only when −1 ≤ c/R ≤ 1. Equivalently |c| ≤ R.

|c| > R: no real roots
|c| = R: one angle family per full turn
|c| < R: two angle families per full turn

Counts in a particular x interval can differ: it may span a fraction of a turn, several turns, or contain both endpoints representing the same circle point. Even |c| < R does not guarantee a root inside a short interval.

Can 3sin x + 4cos x = 6 have a real solution?Worked example

The amplitude is √(3² + 4²) = 5

The expression always lies between −5 and 5.

6 > 5

There is no real solution for any x.

03 / List both periodic branches

Write the families before filtering the transformed interval.

For sin u = k, let β = arcsin k.
u = β + 360°n or u = 180° − β + 360°n

Here n is any integer. In radians the two families are β + 2nπ and π − β + 2nπ. When k = ±1, the two formulae describe the same family; remove duplicates.

For cos u = k, let γ = arccos k.
u = ±γ + 360°n

Recover x only after finding every u in the transformed interval. For m > 0, L ≤ x < U becomes mL + α ≤ u < mU + α. If an input multiplier is negative, reverse the bound order and keep track of which endpoint is included.

Watch the number of intersections change with the target

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Solve an exact shifted sine equation

Included endpoints can both appear in the answer.

Solve √3 sin x + cos x = 1 for 0° ≤ x ≤ 360°.Worked example

√3 sin x + cos x = 2sin(x + 30°)

Match the phase coefficients.

sin(x + 30°) = 1/2

Divide by the amplitude 2.

30° ≤ x + 30° ≤ 390°

Transform the full closed interval.

x + 30° = 30°, 150°, 390°

The upper endpoint is a valid input.

x = 0°, 120°, 360°

Subtract the phase from every angle.

On 0° ≤ x < 360°, remove the last answer. Endpoint conventions are part of the question, not a cosmetic detail.

05 / Solve a signed-coefficient cosine equation

An acute phase can appear with a plus sign in cosine.

Solve 4cos x − 3sin x = 5/2 for 0° ≤ x < 360°.Worked example

4cos x − 3sin x = 5cos(x + β)

Here cos β = 4/5, sin β = 3/5 and β = arctan(3/4) is acute.

cos(x + β) = 1/2

The transformed interval is β ≤ x + β < 360° + β.

x + β = 60°, 300°

Both angles are inside the transformed interval.

x = 60° − β, 300° − β

These are exact degree-angle expressions; evaluate β in degrees if using decimals.

06 / Keep every turn of a multiple angle

The input multiplier can create additional roots.

Solve √3 sin 2x + cos 2x = 1 for 0° ≤ x < 360°.Worked example

2sin(2x + 30°) = 1

The same coefficient matching applies to input 2x.

30° ≤ 2x + 30° < 750°

The transformed interval spans two turns.

2x + 30° = 30°, 150°, 390°, 510°

The value 750° is excluded by the strict upper bound.

x = 0°, 60°, 180°, 240°

Subtract 30°, then divide by 2.

07 / Check a denominator before rearranging

A positive denominator can simplify the domain check.

Solve 1/(4 + √3 sin x + cos x) = 1/5 for 0° ≤ x < 360°.Worked example

The denominator is 4 + 2sin(x + 30°)

It lies between 2 and 6, so never vanishes.

4 + √3 sin x + cos x = 5

Taking reciprocals is valid here.

√3 sin x + cos x = 1

Use the exact shifted sine equation.

x = 0°, 120°

Both satisfy the original nonzero denominator condition.

08 / Reject a candidate from an original zero denominator

Cross-multiplication preserves solutions only on the original domain.

Solve (1 + sin x + cos x)/(1 + sin x) = 2 for 0° ≤ x < 360°.Worked example

Exclude x = 270°

The original denominator is zero there.

cos x − sin x = 1

Cross-multiply only for allowed inputs.

√2 cos(x + 45°) = 1

Use the R formula.

x = 0° or 270° after solving the rearranged equation

These are candidates before the original-domain check.

Only x = 0° is valid

At 270° the original quotient is undefined, not equal to 2.

09 / Squaring changes the candidate set

Restore the original sign condition at the end.

If the original equation is F(x) = c, squaring produces F(x)² = c², which also allows F(x) = −c. Use squaring only when helpful and test all candidates in the original equation.

Compare √3 sin x + cos x = 1 with its squared version on [0°, 360°).Worked example

The original equation gives x = 0°, 120°

It requires the combined wave to equal positive 1.

12 / Recap

Find all branches, then return to the original question.

Match the R formula and check the target against the amplitude.
Transform the whole input interval.
List both periodic families, removing duplicates at tangencies.
Recover every x and respect included or excluded endpoints.
Check original denominators and any sign condition lost through squaring.

Section 1 of 12 · Reduce the equation to one trig function