Solving $$3\cos^2\theta +15\sin2\theta =6.28,$$ I tried all trig identities; it gets too complicated. How can I solve this equation and in general, what should I do to solve this kind of equations?
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2Can you write $sin(2\theta)$ in terms of $cos(\theta)$ using trig identities. You should get a quadratic in $cos(\theta)$. – Paul Mar 21 '20 at 21:50
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1The right side of the given equation, is it $6.28$, or $2\pi$ I think? – Hussain-Alqatari Mar 21 '20 at 21:52
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You can also turn $\cos ^2 \theta$ into $\cos 2\theta$ using the half angle identity, then use the angle sum to group the $\sin$ and $\cos$ terms. – Ross Millikan Mar 21 '20 at 21:53
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@Hussain-Alqatari I checked both with WolframAlpha and 6.28 has moderately ugly solutions, whereas the solution for $2 \pi$ are downright nasty. – ViktorStein Mar 21 '20 at 21:54
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What have you tried? What do you know? Where is this question from? I am voting to close this question due to lack of context. – ViktorStein Mar 21 '20 at 21:54
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Hint
$$3 \cos^2(\theta)+15 \sin(2\theta)=6.28 \\ 15 \sin(2\theta)=6.28-3 \frac{1+\cos(2 \theta)}{2} $$
Square both sides and replace $\sin^2(2 \theta)=1-\cos^2(2 \theta)$. Solve the quadratic.
N. S.
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Avoid squaring as it immediately introduces When do we get extraneous roots?
Divide both sides by $\cos^2\theta$ and write $\tan^2\theta=t$ to find
$$3+30t=6.28(1+t^2)$$ which is a quadratic equation in $t$
Similarly we can divide both sides by $\sin^2\theta$
lab bhattacharjee
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