a. Solve $$ 2 \cos heta=-\sqrt{2} $$ on $$ [0,2 \pi) $$ b. Solve $$ an heta=\sqrt{3} $$ on $$ [0,2 \pi) $$

Question

a. Solve $$ 2 \cos 	heta=-\sqrt{2} $$ on $$ [0,2 \pi) $$ b. Solve $$ 	an 	heta=\sqrt{3} $$ on $$ [0,2 \pi) $$

Vega Bernard · Accepted Answer

For part a: $$ 	heta = \frac{3\pi}{4}, \frac{5\pi}{4} $$ For part b: $$ 	heta = \frac{\pi}{3}, \frac{4\pi}{3} $$

a. Solve \( 2 \cos \theta=-\sqrt{2} \) on \( [0,2 \pi) \) b. Solve \( \tan \theta=\sqrt{3} \) on \( [0,2 \pi) \)