Trigonometry

Verify the identity.

sin2x/cosx + sinx = 2sinx

Solve for all values of x:

6cos2x-7cosx-3=0

If f(x)=cos1/2x-sin2x find the value of
f(pi)

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  1. the first identity is not correct

    LS = sin2x/cosx + sinx
    = 2sinxcosx/cosx + sinx
    = 2sinx + sinx
    = 3 sinx

    You have RS = 2sinx

    6cos2x-7cosx-3=0
    You probably typed this incorrectly and really meant
    6cos^2 x-7cosx-3=0
    if so, let cosx = y, then our equation becomes
    6y^2 - 7y - 3 = 0
    (2y - 3)(3y + 1) = 0
    y = 2/3 or y = -1/3

    cosx = 2/3 or cosx = -1/3

    for cosx = 2/3, x is in I or IV
    x = 48.2 or 311.8°

    for cosx = -1/3, x is in II or III
    x = 109.5 or 250.5°

    for the last one, sub in x = π

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