Trigonometric equations

a) $\sin x=-\frac{1}{2}$. The principal value $\arcsin \left(-\frac{1}{2}\right)=-30^{\circ }$ is not in the interval. The sine is negative in the third and fourth quadrants, and the reference angle is $30^{\circ }$:

$$x=180^{\circ }+30^{\circ }=210^{\circ }\text{and}x=360^{\circ }-30^{\circ }=330^{\circ }$$

b) The principal value is $x_0=\arctan 2\approx 1.107$, in the first quadrant. The tangent has period $\pi$, so

$$x=\arctan 2\text{and}x=\arctan 2+\pi (\approx 4.249)$$

The next one, $\arctan 2+2\pi$, already falls outside the interval.

c) $\cos x=\frac{3}{10}$. The principal value is $x_0=\arccos \frac{3}{10}\approx 1.266$. The cosine is positive in the first and fourth quadrants, and symmetry about the horizontal axis gives $2\pi -x_0$:

$$x=\arccos \frac{3}{10}\text{and}x=2\pi -\arccos \frac{3}{10}(\approx 5.017)$$

a) Let $u=2x$. If $0\le x\le 2\pi$, then $0\le u\le 4\pi$ (two turns). For $\sin u=\frac{\sqrt{3}}{2}$, the principal value is $\frac{\pi }{3}$, and the other solution within one turn is $\pi -\frac{\pi }{3}=\frac{2\pi }{3}$. Adding $2\pi$:

$$u=\frac{\pi }{3},\frac{2\pi }{3},\frac{7\pi }{3},\frac{8\pi }{3}$$

All of them lie in $[0,4\pi ]$. Dividing by $2$:

$$x=\frac{\pi }{6},\frac{\pi }{3},\frac{7\pi }{6},\frac{4\pi }{3}$$

b) Let $u=3x$. If $0\le x\le \pi$, then $0\le u\le 3\pi$ (one and a half turns). For $\cos u=-\frac{1}{2}$: $u=\frac{2\pi }{3}$ and $u=\frac{4\pi }{3}$; adding $2\pi$ gives $\frac{8\pi }{3}\le 3\pi$ (valid) and $\frac{10\pi }{3}>3\pi$ (not valid). Therefore

$$u=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3}⟹x=\frac{2\pi }{9},\frac{4\pi }{9},\frac{8\pi }{9}$$

c) Let $u=x-\frac{\pi }{4}$. If $0\le x\le 2\pi$, then $-\frac{\pi }{4}\le u\le \frac{7\pi }{4}$. For $\tan u=1$: $u=\frac{\pi }{4}$ and $u=\frac{\pi }{4}+\pi =\frac{5\pi }{4}$ (the next one already exceeds $\frac{7\pi }{4}$, and the previous one, $-\frac{3\pi }{4}$, is less than $-\frac{\pi }{4}$). Now $x=u+\frac{\pi }{4}$:

$$x=\frac{\pi }{2}\text{and}x=\frac{3\pi }{2}$$

a) 1. One single ratio: ${\sin }^{2}x=1-{\cos }^{2}x$, so $$2(1-{\cos }^{2}x)-\cos x=1⟹2{\cos }^{2}x+\cos x-1=0$$

2. Factorise with $t=\cos x$: $2t^2+t-1=(2t-1)(t+1)=0$, so $t=\frac{1}{2}$ or $t=-1$. Both values lie in $[-1,1]$.

3. $\cos x=\frac{1}{2}$ gives $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$; $\cos x=-1$ gives $x=\pi$.

b) 1. Double angle: $2\sin x\cos x=\cos x$.

2. Bring everything to one side and factorise (without dividing): $\cos x(2\sin x-1)=0$.

3. $\cos x=0$ gives $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$; $\sin x=\frac{1}{2}$ gives $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$.

c) 1. The equation is not defined when $\cos x=0$ (that is, at $x=\frac{\pi }{2}$ and $\frac{3\pi }{2}$); those values cannot be solutions.

2. With $\tan x=\frac{\sin x}{\cos x}$: $\frac{\sin x}{\cos x}=2\sin x$, so $\sin x-2\sin x\cos x=0$ (we multiply by $\cos x\ne 0$) and

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

3. $\sin x=0$ gives $x=0,\pi ,2\pi$; $\cos x=\frac{1}{2}$ gives $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$. None of them makes $\cos x$ zero, so all five are valid.

In the exam: marks are awarded for reducing to a single ratio or factorising, solving each factor, and giving all the solutions in the interval. You cannot divide by $\cos x$ in b), and in c) remember that $x=0$ and $x=2\pi$ belong to the closed interval.