Best Answer

For simplicity's sake, X represent theta.

This is the original problem: sin2x+ cosX = cos2X + sinX

This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1

With this basic property, you can figure out that

sin2 x=1-cos2x

and

cos2x= 1-sin2x

So we can change the original problem to:

1-cos2x+cosx = 1-sin2X + sinX

-cos2x + cosx =-sin2x + sinX

Basic logic tells you that one of two things are happening.

sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1.

The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4.

If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is:

x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)

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Q: How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?

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Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.

It also equals 13 12.

Cosine squared theta = 1 + Sine squared theta

To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)

The question contains an expression but not an equation. An expression cannot be solved.

1

Yes, it is.

Solve using the quadratic formula

(x-3)(x-2)

X=1

That factors to (a + 1)(a + b) a = -1, -b b = -a

(x - 3)(x + 5)

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