41. -/2 points details my notes larpcalclimaga8 5.4.070. 0/8 submission find the solution(s) of the equation…

Explanation:

Step1: Use sum - to - product formula for cosines

The sum - to - product formula for $\cos A-\cos B=- 2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)$. Here $A=x+\frac{7\pi}{6}$ and $B=x - \frac{7\pi}{6}$. Then $4\cos\left(x+\frac{7\pi}{6}\right)-4\cos\left(x - \frac{7\pi}{6}\right)=-8\sin x\sin\frac{7\pi}{6}$. Since $\sin\frac{7\pi}{6}=-\frac{1}{2}$, the left - hand side of the equation $4\cos\left(x+\frac{7\pi}{6}\right)-4\cos\left(x - \frac{7\pi}{6}\right)$ becomes $4\sin x$. The equation $4\cos\left(x+\frac{7\pi}{6}\right)-4\cos\left(x - \frac{7\pi}{6}\right)=-4$ is equivalent to $4\sin x=-4$, so $\sin x=-1$.

Step2: Solve for $x$ in the given interval

We know that $\sin x=-1$ when $x=\frac{3\pi}{2}$ in the interval $[0,2\pi)$.

Answer:

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

Explanation for second problem:

Step1: Use trigonometric identities

First, use the identity $\sin\left(x+\frac{\pi}{2}\right)=\cos x$ and $\tan(\pi - x)=-\tan x$. The equation $12\sin\left(x+\frac{\pi}{2}\right)+18\tan(\pi - x)=0$ becomes $12\cos x-18\tan x = 0$. Since $\tan x=\frac{\sin x}{\cos x}$, the equation can be rewritten as $12\cos x-18\frac{\sin x}{\cos x}=0$. Multiply through by $\cos x$ (assuming $\cos x\neq0$) to get $12\cos^{2}x - 18\sin x=0$. Use the identity $\cos^{2}x = 1-\sin^{2}x$, then $12(1 - \sin^{2}x)-18\sin x=0$. Expand to get $12-12\sin^{2}x-18\sin x = 0$. Let $t = \sin x$, so $12t^{2}+18t - 12 = 0$. Divide by 6: $2t^{2}+3t - 2=0$.

Step2: Solve the quadratic equation for $t$

Factor the quadratic equation $2t^{2}+3t - 2=(2t - 1)(t + 2)=0$. So $t=\sin x=\frac{1}{2}$ or $\sin x=-2$. Since $-1\leqslant\sin x\leqslant1$, we discard $\sin x=-2$. When $\sin x=\frac{1}{2}$, $x=\frac{\pi}{6}$ or $x=\frac{5\pi}{6}$ in the interval $[0,2\pi)$.

Answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$