Question

Find all solutions of the equation in the interval , 02π. =4cosx+−sin2x4 Write your answer in radians in terms of π. If there is more than one solution, separate them with commas.

Answer 1

The answer is "2nπ".

Explanation:

Given:

`4 \cos x= -\sin^2x+4.......(1)`

We know:

`\Rightarrow \sin^2 x+\cos^2 x=1\\\\\Rightarrow \sin^2 x= 1 -\cos^2 x\\`

put the value of
`\sin^2 x` value in the above equation:

`\Rightarrow 4 \cos x= - (1-\cos^2 x)+4\\\\\Rightarrow 4 \cos x= - 1+\cos^2 x+4\\\\\Rightarrow 4 \cos x= \cos^2 x+3\\\\\Rightarrow \cos^2 x-4 \cos x+3=0\\\\`

Let
`\cos x= A`

`\Rightarrow A^2-4A+3=0 \\ \Rightarrow A^2-(3A+A)+3=0 \\\Rightarrow A^2-3A-A+3=0\\\Rightarrow A(A-3)-1(A-3)=0\\\Rightarrow (A-3)(A-1)=0 \\`

`\Rightarrow A- 3=0 \ \ \ \ \ \ \ \ \ \ \ \Rightarrow A -1 =0 \\\\`

`\Rightarrow A= 3\ \ \ \ \ \ \ \ \ \ \ \Rightarrow A =1 \\\\\Rightarrow \cos x = 3\ \ \ \ \ \ \ \Rightarrow \cos x =1\\\\\Rightarrow x = \cos^(-1) 3\ \ \ \ \ \ \ \Rightarrow \cos x =\cos 0\\\\\Rightarrow x = \cos^(-1) 3\ \ \ \ \ \ \ \Rightarrow x = 0\\\\`

The value of x is
`2n\pi\ \ \ _(where) \ \ \ \ \ \ \ n=1, 2, 3......`

`\boxed{\bold{x=2 n \pi}}`