Homogeneous quadratic trigonometric equations

Content updated: October 22, 2026

An equation involving $\sin^2x,\sin x\cos x,\cos^2x$ has two natural approaches: use power-reduction formulas with angle $2x$, or divide by $\cos^2x$ after checking the case where cosine equals $0$. Both methods must produce the same solution set.

Learning goals: use and cross-check both methods; retain solutions with zero cosine; handle reduced-degree equations and identities in tangent.

Prerequisites: power-reduction formulas, linear equations in sine and cosine, and the domain of tangent.

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

1. The equation form

Consider $a\sin^2x+b\sin x\cos x+c\cos^2x=d$ with real coefficients. Since $1=\sin^2x+\cos^2x$, the equivalent equation is

The left-hand side is homogeneous of degree two in sine and cosine. Coefficients may equal $0$, so the transformed equation may have lower degree or become an identity.

2. Two methods with every case retained

For power reduction, use $\sin^2x=(1-\cos2x)/2$, $\cos^2x=(1+\cos2x)/2$, and $\sin x\cos x=\sin2x/2$. The equation becomes

This is linear in sin 2x and cos 2x. If both coefficients on the left equal $0$, examine the constant on the right instead of dividing by an amplitude of $0$. When solving for $2x$, divide the period as well when recovering $x$.

For the tangent method, first consider $\cos x=0$, or $x=\pi/2+k\pi$. Then $\sin^2x=1$, so this entire family consists of solutions precisely when $a=d$. Next, on the domain $\cos x\ne0$, divide by $\cos^2x$ and set $t=\tan x$ to obtain

If $a=d$, this is linear, an identity, or a contradiction, depending on $b,c-d$. Each finite real root $t_0$ gives $x=\arctan t_0+k\pi$. Finally, take the union with the checked family where cosine equals $0$. If the polynomial in $t$ vanishes identically, still check the branch where cosine equals $0$ before concluding on the entire real line.

3. Worked examples

Example 1. Power reduction and a tangent substitution

Solve $2\sin^2x+\sin x\cos x+3\cos^2x-2=0$. Power reduction followed by multiplication by $2$ yields $\sin2x+\cos2x+1=0$, equivalently $\sin(2x+\pi/4)=-\sqrt2/2$. Dividing by $2$ in the two branches gives

To check by the tangent method, for $\cos x=0$ the left-hand side is $2-2=0$, so accept the family $\pi/2+k\pi$.

On the domain where cosine differs from $0$, divide by cos² and use the identity

The equation becomes

Thus this branch has solutions $x=-\pi/4+k\pi$. The union of the branches agrees exactly with the power-reduction result. Dividing at the outset would lose the family where cosine equals $0$.

Example 2. An identity

Consider $\sin^2x+\cos^2x=1$. This is an identity, so every real number is a solution. The tangent method produces $0=0$ where cosine differs from $0$; the branch where cosine equals $0$ also satisfies the equation. Both domains are needed to conclude that the solution set is all of $\mathbb R$.

4. Practice exercises

1. Solve $\sin^2x-\cos^2x=0$.
2. Solve $\sin^2x+\sin x\cos x=0$.
3. Solve $\sin^2x+\cos^2x=2$.

5. Common mistakes

Divide by cos², not merely cosine, to obtain the quadratic polynomial. Test the case where cosine equals $0$ before dividing. Do not assume the tangent equation is always quadratic. In power reduction, retain the correct sign of cos 2x and divide the period as well when recovering x.

6. Related lessons

• Basic trigonometric equations (English lesson)
• Trigonometric identities (English lesson)
• The equation a sin x + b cos x = c (English lesson)
• Quadratic equations in one trigonometric function (English lesson)
• Higher-degree homogeneous trigonometric equations (English lesson)

7. Video and reference resources

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