Solving trigonometric equationsEdexcel International A Level Maths: Revision notes

A calculator gives only the principal value of an inverse function ($\sin^{-1}$ in $[-90^\circ,90^\circ]$, $\cos^{-1}$ in $[0^\circ,180^\circ]$, $\tan^{-1}$ in $(-90^\circ,90^\circ)$). Use the symmetry of the graphs, or the quadrants, to find the others. If $\alpha$ is the principal value:

• $\sin x=k$: $x=\alpha$ or $180^\circ-\alpha$ (radians $\pi-\alpha$), then add multiples of $360^\circ$ ($2\pi$).
• $\cos x=k$: $x=\pm\alpha$, then add multiples of $360^\circ$.
• $\tan x=k$: $x=\alpha$, then add multiples of $180^\circ$ ($\pi$). A value of $k$ outside $[-1,1]$ has no solution for sine or cosine.

Find the principal value, then list every solution of the form above that lies in the stated interval, and discard the rest. Example: $\sin\theta=0.6$ for $0^\circ\le\theta\le360^\circ$: $\theta=36.9^\circ$ and $143.1^\circ$. These two add to $180^\circ$ because the second is $180^\circ-$ the first. Give answers to the accuracy asked for (for example 1 d.p. in degrees, 3 s.f. in radians) and use exact multiples of $\pi$ when the value is a standard angle such as $\cos x=-\frac12$, giving $x=\frac{2\pi}3$ and $\frac{4\pi}3$ in $0\le x<2\pi$.

When the argument is $x+c$, first transform the interval for $x$ into an interval for $x+c$, solve for the whole argument, then subtract $c$. Example: $\cos(x+30^\circ)=\frac12$, $-180^\circ<x<180^\circ$. Then $-150^\circ<x+30^\circ<210^\circ$. Cosine is $\frac12$ at $\pm60^\circ$, so $x+30^\circ=60^\circ$ or $-60^\circ$, giving $x=30^\circ$ or $-90^\circ$. The value $300^\circ$ is outside the range. Radian example: $\sin\left(x+\frac\pi2\right)=\frac34$ for $0<x<2\pi$. The argument lies in $\left(\frac\pi2,\frac{5\pi}2\right)$ and $\sin^{-1}\frac34=0.848$, so the argument is $\pi-0.848=2.29$ or $2\pi+0.848=7.13$. Then $x=0.723$ or $5.56$.

For $\tan kx=c$ the interval for $x$ is stretched. Multiply the interval by $k$, list the solutions for $kx$, then divide by $k$. Example: $\tan2x=1$ for $90^\circ<x<270^\circ$. Then $180^\circ<2x<540^\circ$. $\tan y=1$ at $y=45^\circ+180^\circ n$, so in range $2x=225^\circ$ or $405^\circ$ and $x=112.5^\circ$ or $202.5^\circ$. A multiple angle usually gives more solutions: $\tan2x=1$ for $0\le x<2\pi$ has four.

Radian solutions are given as multiples of $\pi$ for standard angles ($\frac\pi6,\frac\pi4,\frac\pi3$). For an equation with a square, take the square root and include both signs. Example: $\sin^2\left(x+\frac\pi6\right)=\frac12$ for $-\pi\le x<\pi$. Then $\sin\left(x+\frac\pi6\right)=\pm\frac1{\sqrt2}$ with $-\frac{5\pi}6\le x+\frac\pi6<\frac{7\pi}6$. The solutions for the argument are $-\frac{3\pi}4,-\frac\pi4,\frac\pi4,\frac{3\pi}4$, so $x=-\frac{11\pi}{12},-\frac{5\pi}{12},\frac\pi{12},\frac{7\pi}{12}$.

Use $\sin^2x+\cos^2x=1$ to write an equation in one function, then factorise or use the formula as for any quadratic. Example: $6\cos^2x+\sin x-5=0$ becomes $6\left(1-\sin^2x\right)+\sin x-5=0$, so $6\sin^2x-\sin x-1=0$, or $(3\sin x+1)(2\sin x-1)=0$. Then $\sin x=\frac12$ gives $30^\circ,150^\circ$ and $\sin x=-\frac13$ gives $199.5^\circ,340.5^\circ$. Reject any value of $\sin x$ or $\cos x$ outside $[-1,1]$. Never divide an equation by $\sin x$ or $\cos x$: this loses solutions where that function is zero. Factorise instead, as in $\cos\theta(2\cos\theta-1)=0$.