Question

Find a numerical value of one trigonometric function of x for cos^2x+2sinx-2=0

Answer 1

Explanation:

From

\cos^2x+2\sin x-2=0

Using the identity, we have: \sin^2x+\cos^2x=1 \implying \cos^2x=1-\sin^2x

Opperating:

1-\sin^2x+2\sin x-2=0

-\sin^2x+2\sin x-1=0

\sin^2x-2\sin x+1=0

(\sin x-1)^2=0

\sin x-1=0

\sin x=1

A numerical value for x would be for example x=90 degrees or pi/2 (radians)

And this answer is valid for every angle x=90+360n (n=0,1,2,3,etc) or x=pi/2+2pi*n (n=0,1,2,3,etc)

Answer 2

`\sin x=1`

Explanation:

The given function is

`\cos^2x+2\sin x-2=0`

We use the identity:
`\sin^2x+\cos^2x=1`
`\implies \cos^2x=1-\sin^2x`

This implies that:

`1-\sin^2x+2\sin x-2=0`

`-\sin^2x+2\sin x-1=0`

`\sin^2x-2\sin x+1=0`

`(\sin x-1)^2=0`

`\sin x-1=0`

`\sin x=1`

Hence the numerical value of one trigonometric function(the sine function) is 1