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how do you solve for $0 < \theta < 360$:

$$20\cot\theta + 15\cot\theta\operatorname{cosec}\theta - 4\operatorname{cosec}\theta = 3(1 + \cot^2 \theta) ?$$

I tried turning the $(1 + \cot^2 \theta)$ part into $\operatorname{cosec}^2$ then dividing through or changing every term into $\sin$ and $\cos$ but I always end up with a dodgy equation that I can't factorise.

Michael Hardy
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kjhg
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  • The reason for the conspicuous lack of proper spacing before and after "cosec" is that you used \mbox instead of \operatorname. Contrast these: $$ 15\cot\theta\mbox{cosec}\theta \quad\text{versus}\quad 15\cot\theta\operatorname{cosec}\theta $$ However, you can also use \csc, thus: $$ 15\cot\theta\csc\theta $$ – Michael Hardy Jul 15 '16 at 04:38

1 Answers1

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Let try to convert the equation to $\sin$ and $\cos$.

$$20\frac{\cos \theta}{\sin\theta} + 15 \frac{\cos \theta}{\sin\theta}\frac{1}{\sin\theta} - 4\frac{1}{\sin\theta} = 3\frac{1}{\sin^2\theta}$$ $$20\cos\theta\sin\theta + 15\cos\theta-4\sin\theta = 3$$ $$(4\sin\theta+3)(5\cos\theta-1) = 0$$

Now, you can solve the equation.

GAVD
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