Question

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

Answer 1

Answer:

$\sin x=1$

Step-by-step 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

Answer 2

Answer:

Step-by-step 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)