Wednesday
March 29, 2017

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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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