A quadratic equation in a trigonometric ratio has the form

where $T$ stands for $\sin x$, $\cos x$, or $\tan x$. For example, $5\sin^2 x + 3\sin x - 2 = 0$ is quadratic in $\sin x$ because it can be written as $5A^2 + 3A - 2 = 0$ with $A = \sin x$. The same idea applies to $\cos x$ and $\tan x$.

Solving the quadratic part

Treat the trigonometric ratio as a single unknown and solve the quadratic by the usual methods:

• Factorising when the expression factorises neatly, e.g. $(5\sin x - 2)(\sin x + 1) = 0$.
• The quadratic formula when factorising is not straightforward.
• Completing the square as an alternative to the formula.

Setting each factor equal to zero produces two linear equations in the trigonometric ratio, such as $5\sin x - 2 = 0$ and $\sin x + 1 = 0$.

Solving the resulting trigonometric equations

Each linear equation is solved separately. Rearrange to isolate the ratio, then find the angles whose sine, cosine, or tangent equals that value.

• Use the inverse trigonometric function to obtain the principal value.
• Use the symmetry and periodicity of the trigonometric functions to find all other solutions in the required interval.
• For $\sin x$ and $\cos x$, each equation typically gives two sets of solutions per full turn; for $\tan x$, solutions repeat every $180^\circ$ (or $\pi$ radians).

Points to watch

• Check that every value obtained for $\sin x$ or $\cos x$ lies between $-1$ and $1$; values outside this range give no solutions.
• Combine the solution sets from both factors, and remove any duplicates.
• Always match the answer to the interval and the angle unit (degrees or radians) specified in the question.