General solutions of trigonometric equations

First find the reference angle with a positive ratio on your calculator, then use CAST to place it in the correct quadrants.

• $\sin x=c$: $x=\theta +k\cdot 360°$ or $x=180°-\theta +k\cdot 360°$
• $\cos x=c$: $x=\pm \theta +k\cdot 360°$
• $\tan x=c$: $x=\theta +k\cdot 180°$

Here $c$ is the value of the ratio and $\theta$ is the angle your calculator gives for it (for a negative $c$, use the reference angle and CAST as below).

where $k\in \mathbb{Z}$.

Example. $2\cos x+1=0$, so $\cos x=-\frac{1}{2}$. The reference angle is 60°, and cosine is negative in quadrants 2 and 3:

$x=\pm 120°+k\cdot 360°,k\in \mathbb{Z}$

Always write "$k\in \mathbb{Z}$". It earns a mark.

Substitute integer values of $k$ into the general solution and keep the answers inside the interval.

Example. Solve $\tan (2x-10°)=1,5$ for $0°\le x\le 180°$.

$2x-10°=56,31°+k\cdot 180°$

$x=33,15°+k\cdot 90°$

$k=0$ gives $33,15°$ and $k=1$ gives $123,15°$. $k=2$ gives $213,15°$, which is outside the interval.

Notice the period: dividing by 2 also divides the $k\cdot 180°$ by 2. Forgetting this is the most common error with multiple angles.

If the equation has double angles or mixed ratios, rewrite it with one ratio of one angle, then factorise.

Example. Solve $\sin 2x=\cos x$.

$2\sin x\cos x-\cos x=0$

$\cos x(2\sin x-1)=0$

• $\cos x=0$: $x=90°+k\cdot 180°$
• $\sin x=\frac{1}{2}$: $x=30°+k\cdot 360°$ or $x=150°+k\cdot 360°$

Never divide both sides by a trig ratio such as $\cos x$. You lose the solutions where that ratio is 0. Factorise it out instead.

Equations that are quadratic in a ratio, such as $2{\sin }^{2}x-\sin x-1=0$, factorise like ordinary quadratics. Reject any factor giving a value outside $[-1;1]$ for sine or cosine.