How do you solve the equation $3 sin 2 θ - 2 cos 2 θ = 2$ $3sin2theta - 2cos2theta = 2$ for $0 \leq θ \leq 2 π$ $0 <= theta <= 2pi$ , using the substitution $t = tan θ$ $t=tantheta$ ?

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Answer 1 · 2018-05-20T09:40:08

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

Recall that, if, $tan θ = t$ $tantheta=t$ , then,

$sin 2 θ = \frac{2 tan θ}{1 + {tan}^{2} θ} = \frac{2 t}{1 + t^{2}}$ $sin2theta=(2tantheta)/(1+tan^2theta)=(2t)/(1+t^2)$

and, $cos 2 θ = \frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ} = \frac{1 - t^{2}}{1 + t^{2}}$ $cos2theta=(1-tan^2theta)/(1+tan^2theta)=(1-t^2)/(1+t^2)$ .

Hence, sub.ing in the given eqn. : $3 sin 2 θ - 2 cos 2 θ = 2$ $3sin2theta-2cos2theta=2$ ,

$3 {\frac{2 t}{1 + t^{2}}} - 2 {\frac{1 - t^{2}}{1 + t^{2}}} = 2$ $3{(2t)/(1+t^2)}-2{(1-t^2)/(1+t^2)}=2$ .

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

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

$:. tantheta=2/3$ .

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

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

How do you solve the equation 3sin2theta - 2cos2theta = 23sin2θ−2cos2θ=23sin2theta - 2cos2theta = 2 for 0 <= theta <= 2pi0≤θ≤2π0 <= theta <= 2pi, using the substitution t=tanthetat=tanθt=tantheta?