Solve the equation sin^2x= 3 cos^2x. The value of x that satisfies the equation if x lies in the second quadrant is what degrees?

The value of x that satisfies the equation if x lies in the third quadrant is what degrees? ​

Step-by-step explanation:

sin^2(x)= 3 cos^2(x)

Subtract sin^2(x) on both sides

0=3cos^2(x)-sin^2(x)

By Pythagorean Identity sin^2(x)+cos^2(x)=1, we can replace cos^2(x) with 1-sin^2(x)

0=3(1-sin^2(x))-sin^2(x)

Distribute

0=3-3sin^2(x)-sin^2(x)

Combine like terms

0=3-4sin^2(x)

Add 4sin^2(x) on both sides

4sin^2(x)=3

Divide both sides by 4

sin^2(x)=3/4

Take the square root of both sides

sin(x)=+ or - sqrt(3/4)

sin(x)=+ or - sqrt(3)/sqrt(4)

sin(x)=+ or - sqrt(3)/2

This happens on the unit circle at

60 degrees

120 degrees

240 degrees

300 degrees

If we go around and around again forwards or backwards, we get the infinitely many other solutions including the one's already stated in the following solution sets

(60 +360n) degrees

(120 +360n)degrees

(240 +360n)degrees

(300 +360n)degrees

where n is an integer

x is 120 degrees or (120+360n) degrees in the second quadrant. I didn't know if you were just looking for solutions in the first nonnegative cycle of the unit circle or not.

240 degree one or the (240+360n) degrees terminates in the third.

the answer is 8/21

hope this