Math - Solving Trig Equations

Solve each equation for 0 is less than and equal to "x" is less than and equal to 360

3sinx = 2cos^{2}x

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I don't know how to solve this equation...this is what I have, but I don't know if I'm on the right track or not

3sinx = 1 - 2sin^{2}x
3sinx - 1 + 2sin^{2}x = 0
2sin^{2}x + 3sinx - 1 = 0
sinx(3 + 2sinx) = 1

...and if I am, what do I do next now?

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The answers are:
30 and 150 degrees

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  1. 3sinx = 2cos^2(x)

    I believe you've treated this as cos(2x) and expanded it as a double-angle identity. However, you can't do this because the cosine function is being squared. The 2 is not being multiplied by the x.

    I'm working on it...

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  2. Start by recalling the most important identity. My math teacher calls this "the #1 Identity."

    sin^2(x) + cos^2(x) = 1

    We want to simplify our trig equation by writing everything in terms of sine. Let's solve the #1 Identity for cos^2(x) because we have that in our trig equation.

    cos^2(x) = 1 - sin^2(x)

    Plug that into the trig equation, and see what you can get from there. Let me know if you get stuck along the way.

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  3. wow I forgot about that identity, thank you!

    But, I'm still stuck...

    I get
    cos^2x + sin^2x = 1

    I can't use any compound identities for this...I don't know how to isolate for either cos or sin...

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  4. The only identity you need it the #1 Identity.

    Plugging in 1 - sin^2(x) for cos^2(x), you should get...

    3sinx = 2(1 - sin^2(x))

    Use the distributive property, collect everything on the left, and then factor.

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  5. The only identity you need is the #1 Identity.

    Plugging in 1 - sin^2(x) for cos^2(x), you should get...

    3sinx = 2(1 - sin^2(x))

    Use the distributive property, collect everything on the left, and then factor.

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  6. thanks, I solved it :)

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  7. Glad to be of service. :)

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  8. Is d answer 90degree

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