# Solve cos^2(x)=sin(x) First we must get everything in the same terms. Here I'll choose sin(X) because it is the easiest. So we need to use a trig identity to substitute for `cos^2(X).`

`Cos^2(X)=sin(X)`

`1-sin^2(X)=sin(X)`

Then we can treat this like a quadratic equation. So we will move all the terms to one side and have our equation equal to 0.

`1-sin^2(X)-sin(X)=0`

Now we can rearrange the equation so that it is in standard quadratic equation form. But instead of having the equation in terms of X, we have an equation in terms of sin(X).

`-sin^2(X)-sin(X)+1=0`

`-1(sin(X))^2-1(sin(X))+1=0`

Next, just as we could solve for x in a quadratic equation, we can solve for sin(X) in this equation using the quadratic formula.

`sin(X)=(+1+-sqrt((-1)^2-4(-1)(1)))/(2(-1))`

`sin(X)=(1+-sqrt(5))/-2`

`sin(X)~~-1.618 or 0.618`

As usual with the quadratic formula, we have two possible values.  We will then need to find the degree measure X by finding the arcsine of the values we found above.  But because sine values must fall between -1 and 1, we can throw out -1.618 as a possible value. Thus we only need to use the value 0.618 as our sine value.

`sin(X)~~0.618`

`X~~sin^-1(0.618)`

`X~~38.17^o`

Note that this is not the only degree measure that will give this sign value. There is also a degree measure in Quadrant II that will have the same sine value. We can find that value like so:

`X~~180-38.17=141.83^o`

There are also infinitely many other degree measures that will have this sign value. Every time you move around the unit circle and come upon these degree measures in Quadrants I and II, the sine value will be the same. So all possible values of X are:

`X~~38.17+-360n and X~~141.83+-360n`

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