OMPT trigonometric equations practice: lesson and 10 problems

Answer: 2

Rewriting gives $\sin x=\frac{\sqrt{3}}{2}$. Sine hits that height twice per cycle: once at $x=\frac{\pi }{3}$ and once at the supplementary angle $x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$. So there are 2 solutions.

Answer: A — $\frac{\pi }{6}$

I solve for the inner angle first: $2x=\pm \frac{\pi }{3}+2k\pi$. The smallest positive choice is $2x=\frac{\pi }{3}$, so $x=\frac{\pi }{6}$. Halving at the very end is the whole trick with compressed arguments.

Answer: 3

Factoring gives $\sin x(\sin x-1)=0$, so either $\sin x=0$ or $\sin x=1$. On $[0,2\pi )$, $\sin x=0$ at $x=0$ and $x=\pi$, and $\sin x=1$ at $x=\frac{\pi }{2}$. That is 3 solutions in total.

Answer: B — $x=\frac{\pi }{4}+k\pi$

Tangent has period $\pi$, not $2\pi$, so from the base solution $x=\frac{\pi }{4}$ I step in multiples of $\pi$: $x=\frac{\pi }{4}+k\pi$. The option with $2k\pi$ misses half the solutions, such as $\frac{5\pi }{4}$.

Answer: Step 2

Step 1 is a correct double-angle expansion, and Step 3 solves its equation correctly. The damage happens in Step 2: dividing by $\sin x$ is only legal when $\sin x\ne 0$, and it discards the cases $x=0$ and $x=\pi$. Bringing everything to one side and factoring, $\sin x(2\cos x-1)=0$, gives the full solution set $\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}\}$.

Answer: 6.28

Treating $\cos x$ as the unknown, the quadratic factors as $(2\cos x-1)(\cos x+2)=0$. The root $\cos x=-2$ is impossible, so only $\cos x=\frac{1}{2}$ survives, giving $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$. Their sum is $2\pi \approx 6.28$.

Answer: 120

Isolate first: $\cos \left(\frac{x}{2}\right)=\frac{1}{2}$. Let $u=\frac{x}{2}$. As $x$ runs over $[0,2\pi )$, $u$ only runs over $[0,\pi )$, half a turn. On that half turn $\cos u=\frac{1}{2}$ happens once, at $u=\frac{\pi }{3}$; the usual partner $\frac{5\pi }{3}$ is outside. So $x=2u=\frac{2\pi }{3}$, which is $120^{\circ }$. Check: $2\cos 60^{\circ }=2\cdot \frac{1}{2}=1$.

Answer: A — $x=-\frac{\pi }{3}+2k\pi$ or $x=\frac{4\pi }{3}+2k\pi$

The reference angle is $\frac{\pi }{3}$, and sine is negative in the third and fourth quadrants. Fourth quadrant: $x=-\frac{\pi }{3}$, which is the same point as $\frac{5\pi }{3}$. Third quadrant: $x=\pi +\frac{\pi }{3}=\frac{4\pi }{3}$. Sine repeats every $2\pi$, so both families carry $2k\pi$. The second option lists the positive-sine angles, the third uses the tangent period, and the fourth is the cosine pattern. Check: $\sin \frac{4\pi }{3}=-\sin \frac{\pi }{3}=-\frac{\sqrt{3}}{2}$.

Answer: 300

Two different functions, so I use the Pythagorean identity to reach one of them: $2(1-{\cos }^{2}x)+3\cos x=3$, which rearranges to $2{\cos }^{2}x-3\cos x+1=0$. With $c=\cos x$ this factors as $(2c-1)(c-1)=0$, so $\cos x=\frac{1}{2}$ or $\cos x=1$. The first gives $x=60^{\circ }$ and $x=300^{\circ }$; the second gives $x=0^{\circ }$. Three solutions, and the largest is $300^{\circ }$. Check: $\sin 300^{\circ }=-\frac{\sqrt{3}}{2}$, so $2\cdot \frac{3}{4}+3\cdot \frac{1}{2}=\frac{3}{2}+\frac{3}{2}=3$.

Answer: 4

Isolate: $\sin 3x=\frac{1}{2}$. Let $u=3x$. As $x$ runs over $[0,\pi )$, $u$ runs over $[0,3\pi )$, one and a half turns. Sine equals $\frac{1}{2}$ at $u=\frac{\pi }{6}$ and $\frac{5\pi }{6}$ in the first turn, and at $\frac{13\pi }{6}$ and $\frac{17\pi }{6}$ in the second. Both of those are below $3\pi =\frac{18\pi }{6}$, so all four survive; the next candidate, $\frac{25\pi }{6}$, is out. Dividing by 3 gives $x=\frac{\pi }{18},\frac{5\pi }{18},\frac{13\pi }{18},\frac{17\pi }{18}$, all inside $[0,\pi )$. Four solutions. Check the last: $3\cdot \frac{17\pi }{18}=\frac{17\pi }{6}=2\pi +\frac{5\pi }{6}$, where sine is $\frac{1}{2}$.