Solving trigonometric equationsEdexcel A-Level Maths: Revision notes

Use the inverse function to find the principal value, then use symmetry (or the CAST diagram / graph) to find every solution in the interval: $\sin x=k:\ x=\alpha,\ 180^\circ-\alpha;\qquad\cos x=k:\ x=\alpha,\ 360^\circ-\alpha;\qquad\tan x=k:\ x=\alpha,\ \alpha+180^\circ.$ Then add or subtract full periods ($360^\circ$ for sine and cosine, $180^\circ$ for tangent) until you have all values in the interval. Example: $\sin x=0.5$ for $0\le x\le360^\circ$ gives $30^\circ$ and $150^\circ$. In radians the same rules use $\pi-\alpha$, $2\pi-\alpha$ and $\alpha+\pi$.

For $\sin(x+70^\circ)=0.5$ with $0<x<360^\circ$, first find the interval for the whole bracket: $70^\circ<x+70^\circ<430^\circ$. Solve for $x+70^\circ$ in that interval: $x+70^\circ=150^\circ$ or $390^\circ$ ($30^\circ$ is too small). Then subtract: $x=80^\circ$ or $320^\circ$. Always change the interval first, then solve, then undo the shift last.

For $3+5\cos2x=1$ with $-180^\circ<x<180^\circ$, rearrange to $\cos2x=-\frac25$. The interval for $2x$ is doubled: $-360^\circ<2x<360^\circ$. The principal value is $2x=113.6^\circ$; by symmetry the values are $2x=\pm113.6^\circ$ and $\pm246.4^\circ$ (found using $360^\circ-113.6^\circ$). Halve to give $x=\pm56.8^\circ$ and $\pm123.2^\circ$. For $\sin3x$ in $0\le x<2\pi$ you need $0\le3x<6\pi$, which gives six solutions.

Spot a quadratic when the equation contains $\sin^2x$ or $\cos^2x$. Use $\sin^2x+\cos^2x=1$ to leave one trigonometric function, then factorise or use the quadratic formula treating $\sin x$ (or $\cos x$) as the unknown. Example: $6\cos^2x+\sin x-5=0$ becomes $6\sin^2x-\sin x-1=0$, so $(3\sin x+1)(2\sin x-1)=0$. Then $\sin x=\frac12$ gives $30^\circ$, $150^\circ$ and $\sin x=-\frac13$ gives $199.5^\circ$, $340.5^\circ$. Reject any value outside $-1\le\sin x\le1$, such as $\sin x=2$.

If an equation has $\sin x$ and $\cos x$ in a ratio, divide by $\cos x$ and use the identity. Example: $\sin x=2\cos x$ gives $\tan x=2$, so $x=63.4^\circ$ and $243.4^\circ$ in $0\le x<360^\circ$. In a quadratic such as $2\tan^2x-3\tan x-2=0$, factorise as $(2\tan x+1)(\tan x-2)=0$ and solve each part.

When the question uses radians, the interval is stated in radians and exact answers are multiples of $\pi$: for $2\cos^2x=3-3\sin x$ in $0\le x<2\pi$, $2\sin^2x-3\sin x+1=0$ gives $\sin x=\frac12$ or $1$, so $x=\frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2}$. Check every answer: it must lie in the interval, and substituting it back into the original equation should work. Use the angle mode that matches the question.