Trigonometric Equations

Sol A Dividing both sides of the equation by $\cos\theta$ gives $\sin\theta/\cos\theta=1$ or $\tan\theta=1$. Since $\tan\pi/4=1$ and the tangent has period $\pi$, $\theta=\pi/4+n\pi$ for any integer $n$.

Sol B Dividing both sides of the equation by $\cos\theta$ gives $\sin\theta/\cos\theta=\sqrt{3}$ or $\tan\theta=\sqrt{3}$. Since $\tan\pi/3=\sqrt{3}$ and the tangent has period $\pi$, the solutions are $\theta=\pi/3$ and $\theta=\pi/3+\pi=4\pi/3$.

Sol C Rewriting $\sin2\theta$ as $2\sin\theta\cos\theta$ gives $2\sin\theta\cos\theta=\sin\theta$ or $2\sin\theta\cos\theta-\sin\theta=0$. Factoring yields $\sin\theta(2\cos\theta-1)=0$, so either $\sin\theta=0$ or $\cos\theta=1/2$.

If $\sin\theta=0$, then $\theta=0$ or $\theta=\pi$.

If $\cos\theta=1/2$, then $\theta=\pi/3$ or $\theta=2\pi-\pi/3=5\pi/3$.

Therefore the solutions are $\theta=0, \pi/3, \pi, 5\pi/3$.

Sol 1 Factoring $2\sin^2 \theta - \sin \theta =0$ gives $(\sin \theta)(2\sin \theta -1)=0$, so $\sin\theta=0$ or $\sin \theta =\frac{1}{2}$. If $\sin\theta=0$, then $\theta=0$ or $\theta=\pi$, and if $\sin \theta =\frac{1}{2}$, then $\theta=\frac{\pi}{6}$ or $\theta=\frac{5\pi}{6}$.

Sol 2 Factoring $\cos^2 \theta +\cos \theta =0$ gives $(\cos \theta)(\cos \theta +1)=0$, so $\cos \theta =0$ or $\cos \theta =-1$. If $\cos \theta =0$, then $\theta=\pi/2$ or $\theta=3\pi/2$; and if $\cos \theta =-1$, then $\theta=\pi$.

Sol 3 Squaring both sides of $\cos \theta - \sin \theta =1$ gives $\cos^2 \theta -2\cos \theta \sin \theta +\sin^2 \theta=1$.

Since $\cos^2 \theta +\sin^2 \theta =1$, we have that $1-2\cos \theta \sin \theta=1$; so $-2\cos \theta \sin \theta=0$ and therefore $\cos \theta =0$ or $\sin\theta=0$. If $\cos \theta =0$, then the original equation gives $-\sin \theta =1$ or $\sin \theta=-1$; so $\theta=3\pi/2$.

if $\sin\theta=0$, then the original equation gives $\cos \theta =1$; so $\theta=0$.

Therefore $\theta=0$ and $\theta=3\pi/2$ are the solutions.

Sol 4 $\tan^2 \theta =\tan \theta$ gives $\tan^2 \theta -\tan \theta=0$, so $(\tan\theta)(\tan\theta-1)=0$ and therefore $\tan\theta=0$ or $\tan\theta=1$.

If $\tan\theta=0$, then $\theta=n\pi$, where $n$ is any integer.

If $\tan\theta=1$, then $\theta=\pi/4+n\pi$, where $n$ is any integer.

Sol 5 Substituting $1-\cos^2\theta$ for $\sin^2\theta$ in the equation $2\sin^2 \theta -\cos \theta -1=0$ gives $2(1-\cos^2\theta)-\cos\theta-1=0$ or $2-2\cos^2\theta-\cos\theta-1=0$, so $0=2\cos^2\theta+\cos\theta-1$. Factoring gives $(2\cos\theta-1)(\cos\theta+1)=0$, so $\cos\theta=1/2$ or $\cos \theta =-1$.

If $\cos\theta=1/2$, then $\theta=\pi/3$ or $\theta=2\pi-\pi/3=5\pi/3$;

and if $\cos \theta =-1$, then $\theta=\pi$.

Therefore the solutions are $\theta=\pi/3$,$\theta=5\pi/3$,and $\theta=\pi$.

Sol 6 Since $\cos2\theta=1-2\sin^2\theta$,substituting for $\cos2\theta$ gives $1-2\sin^2\theta=\sin\theta$ or $0=2\sin^2\theta+\sin\theta-1$. Factoring yields $(2\sin\theta-1)(\sin\theta+1)=0$, so $\sin\theta=1/2$ or $\sin \theta=-1$.

If $\sin\theta=1/2$, $\theta=\pi/6$ or $\theta=\pi-\pi/6=5\pi/6$.

If $\sin \theta=-1$, then $\theta=3\pi/2$.

Therefore the solutions are $\theta=\pi/6$, $\theta=5\pi/6$,and $\theta=3\pi/2$.

Sol 7 $2\cos^2 3\theta = \cos 3\theta$ gives $2\cos^2 3\theta-\cos 3\theta=0$, so $(\cos 3\theta)(2\cos 3\theta -1)=0$. Therefore $\cos 3\theta=0$ or $\cos 3\theta=1/2$, so either

$3\theta=\pi/2, 3\pi/2, 5\pi/2, 7\pi/2, 9\pi/2, 11\pi/2$ or

$3\theta=\pi/3, 5\pi/3, 7\pi/3, 11\pi/3, 13\pi/3, 17\pi/3$.

Dividing by 3 to find $\theta$, we obtain the solutions

$\theta=\pi/6,\pi/2,5\pi/6,7\pi/6,3\pi/2,11\pi/6$ and

$\theta=\pi/9,5\pi/9,7\pi/9,11\pi/9,13\pi/9,17\pi/9$.

Sol 8 Since $\cos4\theta=\cos2(2\theta)=1-2\sin^2(2\theta)$, we get the equation $1-2\sin^2(2\theta)=2-3\sin2\theta$; so $0=2\sin^2 2\theta-3\sin2\theta+1$ and therefore $(2\sin2\theta-1)(\sin2\theta-1)=0$.

Then $\sin2\theta=1/2$ or $\sin2\theta=1$, so either

$2\theta=\pi/6, 5\pi/6, 13\pi/6, 17\pi/6$ or

$2\theta=\pi/2, 5\pi/2$.

Dividing by 2 to find $\theta$, we obtain the solutions

$\theta=\pi/12, 5\pi/12, 13\pi/12, 17\pi/12$ and

$\theta=\pi/4, 5\pi/4$.