can't solve a trigo equ

**afeasfaerw23231233** · June 14th 2008, 11:48 PM

i can't solve this trigo equ.

$2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0$
$sin 2xcos^2 x - (1+sinx )(cos2x)= 0$
$tan2x = \frac {1+sinx}{cos^2x}$
?

**red_dog** · June 15th 2008, 12:07 AM

Originally Posted by afeasfaerw23231233

$2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0$

I used the identities $2\sin x\cos x=\sin 2x$ and $\cos^2x-\sin^2x=\cos 2x$
Now, the equation becomes
$\sin 2x\cos x-\cos 2x-\sin x\cos 2x=0\Leftrightarrow \sin x-\cos 2x=0$
(I used the formula $\sin(a-b)=\sin a\cos b-\sin b\cos a$ )
Using the formula $\cos 2x=1-2\sin^2x$ the equation becomes $2\sin^2x+\sin x-1=0$
Let $\sin x=t$ and the equation $2t^2+t-1=0$ has the roots $t_1=-1, \ t_2=\frac{1}{2}$
So you have to solve the equations $\sin x=-1$ and $\sin x=\frac{1}{2}$

**mr fantastic** · June 15th 2008, 12:12 AM

Originally Posted by afeasfaerw23231233

i can't solve this trigo equ.

$2sinx cos^2x - (1+sinx)(cos^2 x - sin ^2 x) = 0$
$sin 2xcos^2 x - (1+sinx )(cos2x)= 0$
$tan2x = \frac {1+sinx}{cos^2x}$
?

Go back to line 1. Substitute $\cos^2 x = 1 - \sin^2 x$ and expand. Let $t = \sin x$ . You end up with (but you better check):

$2t^2 + t - 1 = 0$ .

Solve for t. Then solve $t = \sin x$ for x.

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