Elementary trigonometric equations (II)

Statement

Solve the following equations, drawing the unit circle:

• (a) $2+2\sin(\alpha)=1$
• (b) $-1+5\sin(\alpha)=1$
• (c) $1-\cos(\alpha)=-3\cos(\alpha)$
• (d) $\sin(\alpha)=-\dfrac{\sqrt{3}}{2}$
• (e) $\sin(\alpha)=0$
• (f) $\cos(\alpha)=-\dfrac{1}{2}$

Solution

Isolate the trigonometric function and read the angles on the unit circle (solutions in $[0,2\pi)$, plus the period).

• (a) $\sin\alpha=-\dfrac12 \Rightarrow \alpha=\dfrac{7\pi}{6}+2k\pi \ \vee\ \alpha=\dfrac{11\pi}{6}+2k\pi$.
• (b) $\sin\alpha=\dfrac25=0{.}4$ (not a special value) $\Rightarrow \alpha=\arcsin(0{.}4)\approx0{.}4115+2k\pi \ \vee\ \alpha\approx\pi-0{.}4115=2{.}7301+2k\pi$.
• (c) $1=-2\cos\alpha \Rightarrow \cos\alpha=-\dfrac12 \Rightarrow \alpha=\dfrac{2\pi}{3}+2k\pi \ \vee\ \alpha=\dfrac{4\pi}{3}+2k\pi$.
• (d) $\sin\alpha=-\dfrac{\sqrt3}{2} \Rightarrow \alpha=\dfrac{4\pi}{3}+2k\pi \ \vee\ \alpha=\dfrac{5\pi}{3}+2k\pi$.
• (e) $\sin\alpha=0 \Rightarrow \alpha=k\pi$ (that is $0,\ \pi,\ 2\pi,\dots$).
• (f) $\cos\alpha=-\dfrac12 \Rightarrow \alpha=\dfrac{2\pi}{3}+2k\pi \ \vee\ \alpha=\dfrac{4\pi}{3}+2k\pi$.

$\boxed{\tfrac{7\pi}{6},\tfrac{11\pi}{6};\ \approx0{.}41,\,2{.}73;\ \tfrac{2\pi}{3},\tfrac{4\pi}{3};\ \tfrac{4\pi}{3},\tfrac{5\pi}{3};\ k\pi;\ \tfrac{2\pi}{3},\tfrac{4\pi}{3}}$