1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ How you got $(\sin x + \cos x)(1-\sin x \cos x)$ ??? <It is $\sin ^4x + \cos ^4x$ $\endgroup$ – Aqua Feb 25 '20 at 10:08
  • $\begingroup$ @Aqua Got my mistake. Thanks. $\endgroup$ – Aditya Jain Feb 25 '20 at 10:17
6
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Please replace the coefficient $3$ with $4$ $\endgroup$ – Qurultay Feb 25 '20 at 10:12
  • $\begingroup$ That is great solution. But is there a way to reach answer by following from where I left? $\endgroup$ – Aditya Jain Feb 25 '20 at 10:15
  • $\begingroup$ @Qurultay indeed $\endgroup$ – TZakrevskiy Feb 25 '20 at 10:16
  • 1
    $\begingroup$ @AdityaJain your second formula ("... after rearranging...") seems to be incorrect. $\endgroup$ – TZakrevskiy Feb 25 '20 at 10:19
  • 1
    $\begingroup$ @TZakrevskiy Yeah aqua pointed the mistake and I got the answer. Thanks for help. :) $\endgroup$ – Aditya Jain Feb 25 '20 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.