Question

Find the general solution of the following:

Answer 1

\[\sin (x+ 40) = 1/\sqrt{2}\]
\[\sin5x = \sin 3x\]
\[\tan ax = \cot bx \]
\[\tan ^{2}3x = 3\]

Answer 2

\[x+40 {}^{\circ}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \]\[x=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)-40 {}^{\circ} \]\[x=\frac{\pi }{4}-\frac{\pi (40 {}^{\circ})}{180 {}^{\circ}} \]\[x=\frac{\pi }{36} \]

Answer 3

tan 3x = \[\sqrt{3}\] ,
3x = \[\tan^{-1} \sqrt{3}\]
x=\[ \pi/9\]