
equation using identites
solve the following equation for A, giving all the roots in the interval 0<A<360 (including 0 and 360, i don't know how to write the dash under the sign) correct to the nearest dp:
4((sinA)^2)cosA=(tanA)^2
ok i already know my results are wrong, i don't want the answers i just want to know where i went wrong
so using tan=sinA/cosA, i eliminate (sinA)^2 from both sides of the equation
which gives 4cosA=1/(cosA^2)
=> (cosA^3)=1/4
=> cosA=1/sqrt2
A= 45, 315
where am i wrong?

$\displaystyle \displaystyle 4\sin^2{A}\cos{A} = \tan^2{A}$
$\displaystyle \displaystyle 4\sin^2{A}\cos{A} = \frac{\sin^2{A}}{\cos^2{A}}$
$\displaystyle \displaystyle 4\sin^2{A}\cos^3{A} = \sin^2{A}$
$\displaystyle \displaystyle 4\sin^2{A}\cos^3{A}  \sin^2{A} = 0$
$\displaystyle \displaystyle \sin^2{A}(4\cos^3{A}  1) = 0$
$\displaystyle \displaystyle \sin^2{A} = 0$ or $\displaystyle \displaystyle 4\cos^3{A}  1 = 0$.
Go from here.

Or take roots from here...
The same form is:
Cos[x/2]^2*Csc[[Pi]/4  x/2]^2 Csc[[Pi]/4 + x/2]^2*Sin[x/2]^2 (1 + Cos[3 x]+3*Cos[x]) =0
Best Regards!