Question

I need help with number 7, and the second part of number 8. Thank you!

Answer 1

Answer:

7. $\dfrac{\pi }{3},\ \pi ,\ \dfrac{5\pi }{3}$

8. b. $0,\ \dfrac{2\pi }{3},\ \pi,\ \dfrac{4\pi }{3}$

Step-by-step explanation:

7. Solve the equation

$\cos x+\cos 2x=0$

First, note that

$\cos 2x=2\cos^2x-1$

Hence,

$\cos x+2\cos ^2x-1=0$

Use substitution

$t=\cos x,$

then

$2t^2+t-1=0\\ \\D=1^2-4\cdot 2\cdot (-1)=1+8=9\\ \\\sqrt{D}=\sqrt{9}=3\\ \\t_{1,2}=\dfrac{-1\pm 3}{2\cdot 2}=-1,\ \dfrac{1}{2}$

Now,

$\cos x=-1\ \text{or}\ \cos x=\dfrac{1}{2}$

Solve each equation separately for $0\le x$

$\cos x=-1\\ \\x=\pi+2\pi k, \in Z\\ \\\text{only } x=\pi \in [0,2\pi)$

$\cos x=\dfrac{1}{2}\\ \\x=\pm\arccos \dfrac{1}{2}+2\pi k,\ k\in Z\\ \\x=\dfrac{\pi }{3}\ \text{and}\ x=\dfrac{5\pi }{3}$

8. b. Solve the equation

$\sin x+2\sin x\cos x=0$

Rewrite it:

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

By zero product property,

$\sin x=0\ \text{or}\ 1+2\cos x=0$

Solve each equation separately.

$\sin x=0\\ \\x=\pi k,\ k\in Z\\ \\\text{for}\ x\in [0,2\pi),\ \ \ \ x_1=0\ \ x_2=\pi$

$1+2\cos x=0\\ \\\cos x=-\dfrac{1}{2}\\ \\x=\pm \arccos \left(- \dfrac{1}{2}\right)+2\pi k,\ k\in Z\\ \\x=\dfrac{2\pi }{3}\ \text{and}\ x=\dfrac{4\pi }{3}$