beauscaff
beauscaff
04-05-2017
Mathematics
contestada
cos(2theta)=11 sin(theta)-5
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LammettHash
LammettHash
04-05-2017
[tex]\cos2\theta=11\sin\theta-5[/tex]
[tex]1-2\sin^2\theta=11\sin\theta-5[/tex]
[tex]2\sin^2\theta+11\sin\theta-6=0[/tex]
[tex](2\sin\theta-1)(\sin\theta+6)=0[/tex]
[tex]\implies\begin{cases}2\sin\theta-1=0\\\sin\theta+6=0\end{cases}[/tex]
The second equation gives [tex]\sin\theta=-6[/tex], but [tex]-1\le\sin\varphi\le1[/tex] for any real [tex]\varphi[/tex], so we can ignore this equation.
This leaves us with
[tex]2\sin\theta-1=0\implies \sin\theta=\dfrac12\implies\theta=\dfrac\pi6+2\pi k,\theta=\dfrac{5\pi}6+2\pi k[/tex]
where [tex]k[/tex]is any integer.
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