How do you solve the equation 3sin2theta - 2cos2theta = 23sin2θ2cos2θ=2 for 0 <= theta <= 2pi0θ2π, using the substitution t=tanthetat=tanθ?

1 Answer
May 20, 2018

theta=0.588^c, (3.14+0.588)^c=3.728^c in [0,2pi]θ=0.588c,(3.14+0.588)c=3.728c[0,2π].

Explanation:

Recall that, if, tantheta=ttanθ=t, then,

sin2theta=(2tantheta)/(1+tan^2theta)=(2t)/(1+t^2)sin2θ=2tanθ1+tan2θ=2t1+t2

and, cos2theta=(1-tan^2theta)/(1+tan^2theta)=(1-t^2)/(1+t^2)cos2θ=1tan2θ1+tan2θ=1t21+t2.

Hence, sub.ing in the given eqn. : 3sin2theta-2cos2theta=23sin2θ2cos2θ=2,

3{(2t)/(1+t^2)}-2{(1-t^2)/(1+t^2)}=23{2t1+t2}2{1t21+t2}=2.

:. 6t-2+2t^2=2+2t^2.

;. 6t=4, or, t=4/6=2/3.

:. tantheta=2/3.

:. theta=arctan(2/3) , pi+arctan(2/3).

:. theta=0.588^c, (3.14+0.588)^c=3.728^c in [0,2pi].