# equation using identites

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• Apr 4th 2011, 05:04 AM
furor celtica
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?
• Apr 4th 2011, 05:59 AM
Prove It
$\displaystyle 4\sin^2{A}\cos{A} = \tan^2{A}$

$\displaystyle 4\sin^2{A}\cos{A} = \frac{\sin^2{A}}{\cos^2{A}}$

$\displaystyle 4\sin^2{A}\cos^3{A} = \sin^2{A}$

$\displaystyle 4\sin^2{A}\cos^3{A} - \sin^2{A} = 0$

$\displaystyle \sin^2{A}(4\cos^3{A} - 1) = 0$

$\displaystyle \sin^2{A} = 0$ or $\displaystyle 4\cos^3{A} - 1 = 0$.

Go from here.
• Apr 4th 2011, 06:11 AM
derdack
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!