How Do You Solve a Trigonometric Equation That Reduces to a Quadratic Equation?

Treat one trigonometric ratio as an algebraic variable, solve the resulting quadratic equation, and then convert each valid root back into all angles in the stated finite interval. This is the standard method for IB Mathematics: Analysis and Approaches, SL 3.8: Solving trigonometric equations.

The Method

Suppose you must solve

$$2\sin^2 x-3\sin x+1=0, \qquad 0\leq x\leq 2\pi.$$

First, substitute $u=\sin x$. The equation becomes

$$2u^2-3u+1=0.$$

Factorize and solve:

$$(2u-1)(u-1)=0,$$

so

$$u=\frac{1}{2}\quad\text{or}\quad u=1.$$

Before solving, rewrite the equation so it contains only one trigonometric ratio. If both sine and cosine appear, first use an identity, such as $\sin^2 x+\cos^2 x=1$, to express everything in one ratio. Then make a substitution such as $u=\sin x$, , or . This reveals the quadratic structure and prevents confusion.

Now return to the original trigonometric ratio:

$$\sin x=\frac{1}{2}\quad\text{or}\quad\sin x=1.$$

Use the unit circle to find every solution in the required interval.

Therefore,

$$\boxed{x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}}.$$

After solving the quadratic, reject impossible trigonometric values before applying inverse functions. For example, $u=3/2$ cannot equal sine or cosine because both have range $[-1,1]$. For each value, find a reference angle and use the signs of the ratio in each quadrant. Finally, list answers in increasing order and check them in the original equation.

The same process works when the quadratic is in $\cos x$ or $\tan x$. Check that algebraic roots are possible: $-1\leq\sin x\leq1$ and $-1\leq\cos x\leq1$, while has no equivalent range restriction.

A common misconception is to give only the principal value from an inverse trigonometric function. You must use symmetry and periodicity to find all solutions in the stated interval.

Exam Technique

The command term solve requires relevant algebraic stages, not just answers. On AA Paper 1, show the substitution, factorization and exact unit-circle values; on Paper 2, a GDC may be used, but the equation and interval must still be shown. Do not give a general solution: SL 3.8 requires solutions on a finite interval, and general trigonometric solutions are not required at either level.