Question

solve the equation for 3sin2θ - 2cos2θ = 2 for 0≤θ≤2π using the substitution t= tanθ

Answer 1

\[3*\frac{ 2 \tan \theta }{ 1+\tan ^2\theta }-2*\frac{ 1-\tan ^2\theta }{ 1+\tan ^2 \theta }=2\]
\[6 \tan \theta-2+2\tan ^2\theta =2+2\tan ^2\theta \]
\[6 \tan \theta=2+2=4\]
\[\tan \theta=\frac{ 3 }{ 2 }\]
\[\theta=\tan^{-1} \frac{ 3 }{ 2 },\pi+\tan^{-1} \frac{ 3 }{ 2 }\]