Can I please get some help on these questions:
1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]?
...what about 4cos2θ = 8cos^2 2θ?

2. True or False: sin^2 4x = 1 has 8 solutions on the interval [0, 2pi]?

3. Solve 1 + cos4x = 0 on the interval [0, 2pi]
Is it pi, pi/2, 3pi/2, and 2pi

4. cos^2 4x = 2 has 8 solutions on the interval [0, 2pi]?

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  1. #1 Gotta be careful when solving for 2θ. It increases the possibility for solutions for θ

    2sin^2 2θ = sin2θ
    2sin^2 2θ - sin2θ = 0
    sin2θ(2sin2θ-1) = 0
    sin2θ = 0,1/2

    If sin2θ = 0, 2θ = 0,π,2π,3π,...
    So, π = 0,π/2,π,3π/2,2π

    If sin2θ = 1/2, 2θ = π/6, 5π/6, 13π/6, 17π/6,...
    So, θ = π/12, 5π/12, 13π/12, 17π/12

    Looks like 9 solutions in [0,2π]

    #2 TRUE
    sin^2(4x) = 1
    sin(4x) = ±1
    If sin4x = ±1, 4x is an odd multiple of π/2. There are two of those in [0,2π], so there are in fact 8 solutions for x.

    1+cos4x = 0
    cos4x = -1
    So, 4x is an odd multiple of π. There is just one of those, but when we solve for x rather than 4x, we get 4 solutions. However, they are π/4,3π/4,5π/4,7π/4.

    Plug in your solutions and you will see that cos4x = 1, not -1.

    Since |cos4x| <= 1, cos^2(4x) can never be 2! Trick Question!

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    posted by Steve

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