Posted by **Jon** on Friday, February 15, 2008 at 12:52pm.

2)Find the solution of sin2 theta = cos theta if 0 -< theta < 180

A)30 degrees and 90 degrees

B)30 degrees and 150 degrees

C)30 degrees, 90 degrees, 150 degrees

D)0 degrees, 90 degrees, and 150 degrees

sin2 theta = cos theta

sin2 theta - 1 = 0

(sin theta + 1)(sin theta - 1)= 0

sin theta = -1 or sin theta = 1 which is B

I used the zero property

- Trig(continued) -
**Reiny**, Friday, February 15, 2008 at 1:07pm
For your earlier post for this same question at 3:17 am, I said

"for #2 I too, like Bob, was puzzled

First of all is it

sin 2θ = cos θ or sin^{2} θ = cos θ ?

for the first one

2sinθcosθ - cosθ = 0

cosθ(1sinθ - 1) = 0

cosθ=0 or sinθ=1/2

so θ = 90º or θ = 30º or 150º for your domain, which is choice C "

how did you get your second line???

- Trig(continued) -
**drwls**, Friday, February 15, 2008 at 1:14pm
Please use ^ in front of exponents. Otherwise, we cannot tell sin (2theta) from (sin theta)^2.

Your first step is wrong, since cos theta is not necessarly equal to 1. Also, if sin theta were 1 or -1, why would the answer be B?

I will assume you meant to write sin (2theta) and will let x be theta, to simplify typing.

2 sin x cos x = cos x

2 sin x = 1

sin x = 1/2

x = 30 or 150 degrees.

You seem to have treated sin2theta as sin^2 theta, in your next-to-last step, but if you do that, the answer is completely different

- Trig(continued)Reiny -
**Jon**, Friday, February 15, 2008 at 1:33pm
my question is sin 2è = cos è NOT sin2 è = cos è. I'm sorry, I didn't understand your question earlier.

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