Question

Trigonometry! _/\_

Please answer with proper explanation.

Answer 1

1+ |cosx|+.......,

its infinite gp with common ratio |cosx|

= 1/( 1- |cosx|)

2^2 = 4

compare

1/( 1- |cosx|). = 2

( 1- |cosx|) = 1/2

|cosx| = 1/2

x=(3n-1) pie/3, ( 3n-2) pie/3

Answer 2

Step-by-step explanation:

$Given: 2^{1+|cosx|+cos^2x+|cosx|^3+...\infty} = 4$

$\Rightarrow 2^{1+|cosx|+cos^2x+|cosx|^3+...\infty} = 2^2$

$\Rightarrow 1+|cosx|+cos^2x+|cos^3x|+...\infty} = 2^2$

As we can Clearly see that the given Series is in G.P.

$\Rightarrow S_{n} = \frac{a}{1-r}$

$\Rightarrow \frac{1}{1-|cosx|} = 2$

$\Rightarrow \frac{1}{2} = |1-cosx|$

(i)

$\Rightarrow \frac{1}{2}= |1-cosx|$

$\Rightarrow cosx=\frac{1}{2}$

$\Rightarrow x = \frac{\pi }{3} + 2\pi n$

(ii)

$\Rightarrow \frac{1}{2}= |1-cosx|$

$\Rightarrow cosx=\frac{-1}{2}$

$\Rightarrow cosx =cos \frac{2\pi}{3}$

$\Rightarrow x = 2\pi n + \frac{2\pi}{3}, \ x = 2\pi n + \frac{4\pi}{3}$

The general solutions are:

$\Rightarrow \boxed{ x = \frac{\pi }{3} + 2\pi n, \ x = 2n\pi + \frac{2\pi}{3},x = 2\pi n + \frac{4\pi}{3} }$

Hope it helps!