How do you solve the equation #3sin2theta - 2cos2theta = 2# for #0 <= theta <= 2pi#, using the substitution #t=tantheta#?

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Answer 1 · 2018-05-20T09:40:08

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

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

#sin2theta=(2tantheta)/(1+tan^2theta)=(2t)/(1+t^2)#

and, #cos2theta=(1-tan^2theta)/(1+tan^2theta)=(1-t^2)/(1+t^2)#.

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

#3{(2t)/(1+t^2)}-2{(1-t^2)/(1+t^2)}=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]#.