Solving a sin x + b cos x = c

Content updated: October 22, 2026

The expression $a\sin x+b\cos x$ can be written as a sine wave with amplitude $\sqrt{a^2+b^2}$. This identity determines when the equation has solutions and reduces it to a basic trigonometric equation.

Learning goals: choose an auxiliary angle with the correct signs; find every solution family; handle zero coefficients and boundary values.

Prerequisites: sine and cosine addition formulas and basic trigonometric equations.

On this page: The idea · Method · Examples · Practice · Further learning

1. Amplitude and the degenerate case

Consider $a\sin x+b\cos x=c$, where $a,b,c\in\mathbb R$. If $a=b=0$, the equation becomes $0=c$: every real number is a solution when $c=0$, and there are no solutions when $c\ne0$. Do not divide by $\sqrt{a^2+b^2}$ in this case.

Suppose $R=\sqrt{a^2+b^2}>0$. Since $(a/R)^2+(b/R)^2=1$, an angle $\alpha$ exists such that $\cos\alpha=a/R$ and $\sin\alpha=b/R$. These two conditions together determine its position on the unit circle.

2. Reduction to a sine equation

Since the range of sine is exactly $[-1;1]$, the equation has a solution if and only if $|c|\le R$, equivalently $c^2\le a^2+b^2$. This criterion also holds when $R=0$, as checked separately above; division is used only for $R>0$.

If $|c|\le R$, set $\beta=\arcsin(c/R)$. The two solution branches are

At $c=R$ or $c=-R$, the two branches coincide in one family. Choosing the auxiliary angle from $\tan\alpha=b/a$ alone may lose quadrant information and also requires $a\ne0$. Check both the sine and cosine of the chosen angle. Replacing $\alpha$ by its sum with $2\pi$ leaves the solution set unchanged.

Alternatively, use $R\cos(x-\theta)$ with $\cos\theta=b/R$ and $\sin\theta=a/R$. Both forms are valid; choose one that reveals a special angle, then check it with the addition formula.

3. Worked examples

Example 1. Auxiliary angle π/3

Solve $\sin x+\sqrt3\cos x=1$. Here $R=2$, $\cos(\pi/3)=1/2$, and $\sin(\pi/3)=\sqrt3/2$. Thus the equation is equivalent to $\sin(x+\pi/3)=1/2$. Hence

Substituting a representative angle from either family into the left-hand side gives $1$; periodicity with period $2\pi$ preserves the equality for every $k$.

Example 2. Using the cosine form

Solve $\sin x+\cos x=1$. The addition formula gives $\sqrt2\cos(x-\pi/4)=1$, so $x-\pi/4=\pm\pi/4+2k\pi$. Hence $x=2k\pi$ or $x=\pi/2+2k\pi$.

Example 3. An answer using an inverse function

Solve $\sin x+\sqrt3\cos x=1/2$. Writing the left-hand side as $2\cos(x-\pi/6)$ yields $\cos(x-\pi/6)=1/4$. Set $\gamma=\arccos(1/4)\in[0;\pi]$. The solutions are $x=\pi/6\pm\gamma+2k\pi$. The value $1/4$ belongs to $[-1;1]$, so both branches exist.

4. Practice exercises

1. Solve $\sin x-\cos x=0$.
2. Solve $3\sin x+4\cos x=6$.
3. Discuss the solutions of $0\sin x+0\cos x=c$, then solve $2\sin x=2$.

5. Common mistakes

Do not divide by an amplitude of $0$. Do not change the auxiliary-angle sign arbitrarily: expand and check the coefficients of sine and cosine. Before applying an inverse function, check that its input belongs to $[-1;1]$. Include both branches, then combine them if they coincide.

6. Related lessons

• Basic trigonometric equations (English lesson)
• Trigonometric identities (English lesson)
• Quadratic equations in one trigonometric function (English lesson)
• Equations symmetric in $\sin x$ and $\cos x$ (English lesson)
• Homogeneous quadratic trigonometric equations (English lesson)

7. Resources from the original article

• Scribd exercise resource from original article 1
• Registration link from the original article