Solving trigonometric equationsAQA A-Level Maths: Revision notes

To solve $\sin x=k$, $\cos x=k$ or $\tan x=k$ in a given interval:

1. Use the inverse function to find the principal value $\alpha$ from the calculator.
2. Use the symmetry of the graph (or the quadrants) to find the other solutions in one period.
3. Add or subtract whole periods until you have covered the interval. In degrees, for $0^\circ\le x\le360^\circ$: $\sin x=k$ gives $\alpha$ and $180^\circ-\alpha$; $\cos x=k$ gives $\alpha$ and $360^\circ-\alpha$; $\tan x=k$ gives $\alpha$ and $\alpha+180^\circ$. Example: $\sin x=-0.6$. The reference angle is $36.9^\circ$, and sine is negative in the third and fourth quadrants, so $x=216.9^\circ$ and $323.1^\circ$.

In radians, replace $180^\circ$ by $\pi$ and $360^\circ$ by $2\pi$: $\sin x=k$ gives $\alpha$ and $\pi-\alpha$; $\cos x=k$ gives $\alpha$ and $2\pi-\alpha$; $\tan x=k$ gives $\alpha$ and $\alpha+\pi$. If $k$ is an exact value such as $-\frac12$ or $\sqrt3$, give answers as exact multiples of $\pi$. Example: $\cos x=-\frac12$ for $0\le x\le2\pi$ has reference angle $\frac{\pi}{3}$, and cosine is negative in the second and third quadrants, so $x=\frac{2\pi}{3},\frac{4\pi}{3}$. Check your calculator mode first.

For an equation such as $\sin(2x)=k$ or $\cos(3x-30^\circ)=k$, the solutions are for the whole expression, so change the interval first, solve for the whole angle, then undo the operation. Example: $\sin2x=\frac{\sqrt3}{2}$ for $0^\circ\le x\le180^\circ$. Then $0^\circ\le2x\le360^\circ$, so $2x=60^\circ$ or $120^\circ$, giving $x=30^\circ$ or $60^\circ$. Example: $\cos(3x-30^\circ)=0.5$ for $0^\circ\le x\le180^\circ$. The range for $3x-30^\circ$ is $-30^\circ$ to $510^\circ$, so $3x-30^\circ=60^\circ,300^\circ,420^\circ$, and $x=30^\circ,110^\circ,150^\circ$. A multiple of $n$ in the angle gives about $n$ times as many solutions.

If an equation contains $\sin^2x$ and $\sin x$ (or the same for $\cos$ or $\tan$), treat it as a quadratic in that function: factorise or use the formula, then solve each linear equation. Example: $2\sin^2\theta+\sin\theta-1=0$ factorises as $(2\sin\theta-1)(\sin\theta+1)=0$, so $\sin\theta=\frac12$ or $\sin\theta=-1$, giving $\theta=30^\circ,150^\circ,270^\circ$ in $0^\circ\le\theta\le360^\circ$. Always reject a root outside the range of the function: $\sin x$ and $\cos x$ lie between $-1$ and $1$, so $\cos x=3$ has no solution.

If both $\sin$ and $\cos$ appear, use an identity to leave one function:

• $\sin^2x=1-\cos^2x$ or $\cos^2x=1-\sin^2x$
• $\tan x=\frac{\sin x}{\cos x}$ Example: $2\sin^2x+5\cos x+1=0$. Replace $\sin^2x$ to get $2\cos^2x-5\cos x-3=0$, so $(2\cos x+1)(\cos x-3)=0$ and $\cos x=-\frac12$ (reject $\cos x=3$), giving $x=120^\circ,240^\circ$. Example: $\tan x=3\sin x$ becomes $\sin x=3\sin x\cos x$, so $\sin x(1-3\cos x)=0$ and $\sin x=0$ or $\cos x=\frac13$.