# Trig(continued)

posted by on .

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) - ,

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 sin2 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) - ,

line: cosθ(1sinθ - 1) = 0

should have said

cosθ(2sinθ - 1) = 0

• Trig(continued) - ,

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 - ,

my question is sin 2è = cos è NOT sin2 è = cos è. I'm sorry, I didn't understand your question earlier.

• Trig(continued)Reiny - ,

In that case the solution I just gave you above will be the correct one.

Notice that all 3 answers work in your original equation.

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