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I'm stuck on this question. I've tried looking at online trig calculators and I still don't understand what to do.

Solve the following equation algebraically for $0 ≤ x ≤ 2\pi)$.

$\csc^2x + 2\cot x - 5 = 0$

AAA
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  • To solve such equations, the trick is to use substitution formulas to avoid having different functions. In this particular case, you can express $\csc^2 x$ as a function of $\cot x$. Then setting $t=\cot x$, you get an ordinary (non trigonometric) equation. Have a look at http://math.stackexchange.com/q/1814187/65203 –  Jun 15 '16 at 12:53

4 Answers4

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Use $\csc^2x=1+\cot^2x$. That gives $\cot^2x+2\cot x=4$. Solving the quadratic we get roots $-1\pm\sqrt5$.

almagest
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Multiply everything by $\sin^2x$ to get: $$1+2\cos x \sin x-5\sin^2 x = 0$$ Now rewrite $-5\sin^2 x = -4\sin^2 x+\cos^2x-1$ to get rid of the constant term: $$\color{red}{1}+2\cos x \sin x -4\sin^2 x+\cos^2x\color{red}{-1} = 0 \iff 2\cos x \sin x -4\sin^2 x+\cos^2x = 0$$ Divide everything by $\cos^2 x$ to get: $$-4\tan^2 x+2\tan x + 1 = 0$$ This is a quadratic equation in $\tan x$.

Alternatively, immediately multiply everything by $\tan^2 x$ to get: $$\sec^2 x + 2 \tan x - 5\tan^2 x = 0$$ And use $\sec^2x = 1+\tan^2x$ once to arrive at the same equation as above.

StackTD
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$$\csc^2x+2\cot x=\frac{1+\sin 2x}{\sin^2x}=\frac{(\sin x+\cos x)^2}{\sin^2x}=5$$ we have $$\sin x+\cos x=\sqrt{5}\sin x\implies \tan x=\frac{1}{\sqrt{5}-1}$$ or $$\sin x+\cos x=-\sqrt{5}\sin x\implies \tan x=-\frac{1}{\sqrt{5}+1}$$

Behrouz Maleki
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With $t=\cot(x)$, we have $\csc^2(x)=t^2+1$, and $t^2+2t+1=5.$