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Question: For the equation \(\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x + \pi = 0\), the number of real solution is...

For the equation cos⁡−1x+cos⁡−12x+π=0\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x + \pi = 0, the number of real solution is.

A

1

B

2

C

0

D

∞\infty

Answer

0

Explanation

Solution

cos⁡−12x=−π−cos⁡−1x\cos ^ { - 1 } 2 x = - \pi - \cos ^ { - 1 } x

⇒2x=cos⁡(π+cos⁡−1x)\Rightarrow 2 x = \cos \left( \pi + \cos ^ { - 1 } x \right)

⇒ 2x=cos⁡π(cos⁡cos⁡−1x)−sin⁡πsin⁡(cos⁡−1x)2 x = \cos \pi \left( \cos \cos ^ { - 1 } x \right) - \sin \pi \sin \left( \cos ^ { - 1 } x \right)

2x=−x⇒x=02 x = - x \Rightarrow x = 0

But x=0x = 0 does not satisfy the given equation.

No solution will exist.