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Math Help - Trigonometric equations

  1. #1
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    Trigonometric equations

    Solve for x
    1. sin3x= \frac{3}{4}
    2. 4sinx.cosx.cos2x=1
    3. -cos4x=2
    4. cos(x+\frac{pi}{18})=\frac{2}{5}

    Are there some certain ways to solve trig problems like these? Please help me. Thank you
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  2. #2
    MHF Contributor Unknown008's Avatar
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    For the second one, you need to know one identity in particular.

    sin(2A) = 2sinAcosA

    [4sin(x)cos(x)]cos(2x) = [2sin(2x)]cos(2x) = sin(4x)

    Then, you solve for x.

    For that, you find the first value of x as being the inverse of sine.

    I'll call this angle alpha.

    \alpha = sin^{-1}(1) = 90^o

    Then, you need to know in which quadrant sin is positive. It is positive in the first and second quadrant.

    So, the value of 4x becomes 90 degrees.

    Now, since x has a coefficient of 4, you need to consider 4 cycles, that is, you need to include 90+360, 90+720 and 90+1080.

    This gives:

    4x = 90, 450, 810, 1170 degrees

    So, now you can find the value of x for each of those values. Note that I'm taking x to be within the interval of 0^o \leq x \leq 360^o
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