Answer:
See below for answers
Step-by-step explanation:
Problem 1
[tex]\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}[/tex]
Therefore, the second option is correct.
Problem 2
[tex]cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)[/tex]
[tex]0=sinx\\\\x=\pi n\\\\x=0,\pi[/tex]
[tex]0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}[/tex]
Therefore, the solution set is [tex]x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}[/tex], making the second option correct.
Problem 3
[tex]2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0[/tex]
[tex]cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ[/tex]
[tex]2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ[/tex]
Therefore, the solution set is [tex]x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}[/tex], making the fourth option correct
