Solving Quadratic Trigonometric Equations

Definition Suppose an equation can be written as $A(x) \, B(x) = 0$, where $A$ and $B$ are expressions in trigonometric functions of $x$. At every $x$ where both factors are defined, the product is zero exactly when $A(x) = 0$ or $B(x) = 0$. So the solutions are the solutions of $A(x) = 0$ together with those of $B(x) = 0$, restricted to inputs where both factors are defined. To use this, move every term to one side and factor. Do not divide both sides by a trigonometric expression, because that loses every solution that makes the expression zero.

Definition Suppose an equation can be written as

where $u = T(x)$ is a single trigonometric expression such as $\sin x$ or $\tan 2x$. Solve this quadratic for $u$ by factoring or by the quadratic formula. Discard any root that is not in the range of $T$, and for each remaining root $u_0$ solve the linear equation $T(x) = u_0$. If the equation mixes two functions, an identity such as $\sin^2 x + \cos^2 x = 1$ can often rewrite it in a single function first.

Definition An equation with a multiple angle contains a trigonometric function of $nx$ for a constant $n$ other than $0$ and $1$, as in $\sin 2x$ or $\tan 3x$. If every term uses the same angle $nx$, substitute $\theta = nx$, solve for $\theta$, and divide each solution by $n$. The period of $T(nx)$ is the period of $T$ divided by $\lvert n \rvert$, so with $n = 2$ the solutions come twice as often as for $T(x)$. If the equation mixes angles such as $x$ and $2x$, first use a double-angle formula to write everything in terms of one angle.

Definition Let $x$ be restricted to an interval with endpoints $p < q$, and let $\theta = nx$ with $n > 0$. Then $\theta$ is restricted to the interval with endpoints $np$ and $nq$, with the same endpoints included or excluded. Every solution for $\theta$ in this stretched interval must be found before dividing by $n$.

Definition A candidate is a value found by solving an equation derived from the original one. It is extraneous if it satisfies the derived equation but not the original. Squaring both sides can create extraneous candidates, since $a^2 = b^2$ holds when $a = -b$ as well as when $a = b$. Multiplying both sides by an expression that can be zero can create them too. After either step, every candidate must be checked in the original equation.

Definition A solution must lie in the domain of every expression in the original equation. Tangent and secant require $x \neq \frac{\pi}{2} + k\pi$, and cotangent and cosecant require $x \neq k\pi$, for every integer $k$. The zero product rule only applies where every factor is defined, so after factoring or using identities, each candidate must be checked against these restrictions.