Answer:
(a) [tex]x=\frac{5\pi }{3}[/tex]
(b) [tex]x=\frac{\pi }{8}[/tex]
Step-by-step explanation:
It is given that [tex]0\leq x\leq 2\pi[/tex]
(a) We have given equation [tex]2sinx+\sqrt{3}=0[/tex]
[tex]sinx=\frac{-\sqrt{3}}{2}[/tex]
[tex]x=sin^{-1}(-0.866)[/tex]
[tex]x=\frac{-\pi }{3}=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}[/tex]
(b) [tex]tan(2x)=1[/tex]
[tex]2x=tan^{-1}1[/tex]
[tex]2x=\frac{\pi }{4}[/tex]
[tex]x=\frac{\pi }{8}[/tex]
