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September 17, 2014

September 17, 2014

Posted by **.** on Tuesday, April 20, 2010 at 4:03pm.

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)

- Trigonometry -
**Reiny**, Tuesday, April 20, 2010 at 5:33pmthe 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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