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
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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