tana^2 = sin(2a)
tried everything... -_-
$\displaystyle \tan^2 a = \sin (2a)$
We'll make use of the identity: $\displaystyle \sin (2a) = \frac{2\tan a}{1 + \tan^2 a}$
So:
$\displaystyle \begin{aligned} \tan^2 a & = \frac{2\tan a}{1 + \tan^2 a} \\ \tan^2 a (1 + \tan^2 a) & = 2\tan a \\ & \ \vdots \\ \tan^4 a + \tan^2 a - 2\tan a & = 0 \\ \tan a \left(\tan^3 a + \tan a - 2\right) & = 0 \\ \tan a (\tan a - 1) (\tan^2 a + \tan a + 2) & = 0 \end{aligned}$
Set each factor equal to 0 and solve for $\displaystyle a$. Note that the third factor (essentially a quadratic) does not have real solutions for $\displaystyle \tan a$ so don't worry about it.
Exactly. That's why there are no solutions that are real numbers. There is no angle such that when you plug it into $\displaystyle \tan^2 a + \tan a + 2$ you will get 0.
To prove the identity: $\displaystyle \sin (2a) = \frac{2\tan a}{1+\tan^2a}$
Why don't you try ? It isn't too difficult really. Start off with using: $\displaystyle \sec^2 x = 1 + \tan^2 x$
nah i just said i changed the right side to tan by multiplying top and bottom by cosa^2
so 0,pi/4, pi, 5pi/4 r my solutions since
tana = 0 , a can only equal = {0, pi} in the interval i was given
tana = 1, a can only equal = (5pi/4, pi/4} in the interval between 0 and 2pi noninclusive