Question

solve cos2θ = 3sinθ + 2 for 0 < θ < 2π?

Answer 1

http://imgur.com/0969SXI.jpg
Help would be appreciated :)

Answer 2

\[\cos 2 \theta=3 \sin \theta +2\]
\[1-2 \sin ^2 \theta=3 \sin \theta+2,2 \sin ^2\theta +3 \sin \theta+1=0\]
make factors and solve.

Answer 3

@surjithayer - Thank you! I'm' trying to figure it out, but I'm stuck.. I feel a little dumb but could you please further explain? How would I factor the two? Would I take the square root of sin^2θ first?

Answer 4

\[2 \sin ^2 \theta+2 \sin \theta+\sin \theta+1=0\]
\[2 \sin \theta \left( \sin \theta+1 \right)+1\left( \left( \sin \theta+1 \right) \right)=0\]
\[\left( \sin \theta+1 \right)\left( 2 \sin \theta+1 \right)=0\]
\[\sin \theta=-1=\sin \frac{ 3 \pi }{ 2 },\theta=\frac{ 3 \pi }{ 2 }\]
or \[\sin \theta=-\frac{ 1 }{ 2 }=-\sin \frac{ \pi }{ 6 }=\sin \left( \pi+\frac{ \pi }{ 6 } \right),\sin \left( 2 \pi-\frac{ \pi }{ 6 } \right)\]
\[\theta=\frac{ 7 \pi }{ 6 },\frac{ 11 \pi }{ 6 }\]

Answer 5

Thanks a lot. It really helped.

Answer 6

yw