Answer:
7. [tex]\dfrac{\pi }{3},\ \pi ,\ \dfrac{5\pi }{3}[/tex]
8. b. [tex]0,\ \dfrac{2\pi }{3},\ \pi,\ \dfrac{4\pi }{3}[/tex]
Step-by-step explanation:
7. Solve the equation
[tex]\cos x+\cos 2x=0[/tex]
First, note that
[tex]\cos 2x=2\cos^2x-1[/tex]
Hence,
[tex]\cos x+2\cos ^2x-1=0[/tex]
Use substitution
[tex]t=\cos x,[/tex]
then
[tex]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}[/tex]
Now,
[tex]\cos x=-1\ \text{or}\ \cos x=\dfrac{1}{2}[/tex]
Solve each equation separately for [tex]0\le x<2\pi[/tex]
[tex]\cos x=-1\\ \\x=\pi+2\pi k, \in Z\\ \\\text{only } x=\pi \in [0,2\pi)[/tex]
[tex]\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}[/tex]
8. b. Solve the equation
[tex]\sin x+2\sin x\cos x=0[/tex]
Rewrite it:
[tex]\sin x(1+2\cos x)=0[/tex]
By zero product property,
[tex]\sin x=0\ \text{or}\ 1+2\cos x=0[/tex]
Solve each equation separately.
[tex]\sin x=0\\ \\x=\pi k,\ k\in Z\\ \\\text{for}\ x\in [0,2\pi),\ \ \ \ x_1=0\ \ x_2=\pi[/tex]
[tex]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}[/tex]
