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Verify the identity.
sin2x/cosx + sinx = 2sinx
Solve for all values of x:
6cos2x7cosx3=0
If f(x)=cos1/2xsin2x find the value of
f(pi)

Trigonometry 
Reiny,
the first identity is not correct
LS = sin2x/cosx + sinx
= 2sinxcosx/cosx + sinx
= 2sinx + sinx
= 3 sinx
You have RS = 2sinx
6cos2x7cosx3=0
You probably typed this incorrectly and really meant
6cos^2 x7cosx3=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