General solution of trigonometric equations

Find the reference angle using the positive value of the ratio, then use CAST to choose the 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, $\theta$ is the angle your calculator gives for it, and $k\in \mathbb{Z}$.

with $k\in \mathbb{Z}$, where $\theta$ is the angle from the calculator.

Example. $2\sin x=-\sqrt{3}$, so $\sin x=-\frac{\sqrt{3}}{2}$. Reference angle 60°. Sine is negative in quadrants 3 and 4:

$x=240°+k\cdot 360°\text{or}x=300°+k\cdot 360°,k\in \mathbb{Z}$

Substitute $k=0,\pm 1,\pm 2,\dots$ and keep the answers in the interval.

Example. $3\tan x-2=0$ for $-180°\le x\le 180°$.

$\tan x=\frac{2}{3}$, so $x=33,69°+k\cdot 180°$.

• $k=0$: $33,69°$
• $k=-1$: $-146,31°$

Both are in the interval.

Factorise when there is a common factor or a quadratic form.

$2{\cos }^{2}x-\cos x=0\Rightarrow \cos x(2\cos x-1)=0$

So $\cos x=0$ or $\cos x=\frac{1}{2}$.

Use the square identity to get one ratio:

$2{\sin }^{2}x+3\cos x=3\Rightarrow 2(1-{\cos }^{2}x)+3\cos x-3=0\Rightarrow 2{\cos }^{2}x-3\cos x+1=0$

Use the quotient identity when sine and cosine appear once each:

$\sin x=2\cos x\Rightarrow \tan x=2$

You may divide by $\cos x$ here only because $\cos x=0$ would make $\sin x=\pm 1$, which does not satisfy the equation. Say so in your answer.

Never divide by a ratio that could be 0 and give a solution. Factorise it out instead.