Question

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

Answer 1

7.
`(\pi )/(3),\ \pi ,\ (5\pi )/(3)`

8. b.
`0,\ (2\pi )/(3),\ \pi,\ (4\pi )/(3)`

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\\ \\√(D)=√(9)=3\\ \\t_(1,2)=(-1\pm 3)/(2\cdot 2)=-1,\ (1)/(2)`

Now,

`\cos x=-1\ \text{or}\ \cos x=(1)/(2)`

Solve each equation separately for
`0\le x<2\pi`

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

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