# Solve the following trigonometric equation for x

Solve $$\sin^2x \tan x + \cot x \cos^2 x - \sin2x = 1 + \tan x + \cot x$$

I converted the whole equation in $$\sin x$$ and $$\cos x$$ and after rearranging a bit I got $$(\sin x + \cos x)(1-\sin x \cos x) - 2 \sin^2x \cos ^2x = \sin x \cos x +1$$

I supposed $$\sin x+\cos x$$ to be equal to $$t$$ and hence, $$\sin x\cos x$$ will be equal to $$\frac{(t^2-1)}2$$.

Substituting the above values, I got the following equation

$$t^4+t^3-t^2-t+2=0$$

I am unable to find roots of this equation and have got no other way to proceed.

• How you got $(\sin x + \cos x)(1-\sin x \cos x)$ ??? <It is $\sin ^4x + \cos ^4x$ – Aqua Feb 25 '20 at 10:08
• @Aqua Got my mistake. Thanks. – Aditya Jain Feb 25 '20 at 10:17

## 1 Answer

Let for the sake of writing less stuff $$c=\cos x$$ and $$s = \sin x$$.

Case 1: $$cs \ne0$$; we can multiply your equation by $$cs$$ without losing roots. We get $$s^4 + c^4 - 2 s^2c^2 = sc + s^2+c^2$$ after that $$(s^2+c^2)^2 - 4 s^2c^2 = sc + 1$$ or $$4s^2c^2+sc=0.$$

Since $$sc\ne0$$, we get $$4sc+1=0$$, or, in terms of $$x$$, $$2\sin 2x + 1=0$$, which is easy to solve.

Cases 2: $$cs=0$$ is impossible, because otherwise either $$\tan x$$ or $$\cot x$$ is undefined.

• Please replace the coefficient $3$ with $4$ – Qurultay Feb 25 '20 at 10:12
• That is great solution. But is there a way to reach answer by following from where I left? – Aditya Jain Feb 25 '20 at 10:15
• @Qurultay indeed – TZakrevskiy Feb 25 '20 at 10:16
• @AdityaJain your second formula ("... after rearranging...") seems to be incorrect. – TZakrevskiy Feb 25 '20 at 10:19
• @TZakrevskiy Yeah aqua pointed the mistake and I got the answer. Thanks for help. :) – Aditya Jain Feb 25 '20 at 10:21