Question

Hey guys....answer this one...✌​

Answer 1

Answer:

So the solution is 2π / 3

Step-by-step explanation:

We Have :-

$cos^{-1} (\frac{1}{2}) + 2 sin^{-1} (\frac{1}{2})$

To Find :-

Solution of given equation

Solution :-

$First\ lets\ solve\ sin^{-1}(\frac{1}{2})\\\\Let\ y = sin^{-1}(\frac{1}{2})\\\\sin\ y = \frac{1}{2} \\\\sin\ y = sin(\frac{\pi }{6} )\\\\So\ the\ Principal\ Value\ is\ \frac{\pi }{6} \\\\Lets\ solve\ cos^{-1} (\frac{1}{2}) \\\\Let\ x = cos^{-1} (\frac{1}{2} )\\\\cos\ x = \frac{1}{2} \\\\cos\ x = cos(\frac{\pi }{3} )\\\\So\ the\ principa\ value\ is\ \frac{\pi}{3} \\\\Solving\\\\cos^{-1} (\frac{1}{2})+2sin^{-1}(\frac{1}{2})\\\\= \frac{\pi}{3}+2*\frac{\pi}{6}$

$= \frac{\pi}{3} + \frac{\pi}{3}\\\\= \frac{2\pi }{3}$

So the solution is 2π / 3