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Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$: $$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$

This is what I did: $$\begin{array}{crcl} \Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\ \Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\ \Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\ \Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\ \Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$

So, what's next?! Thank you in advance!

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Just found :… – lab bhattacharjee Jan 29 '14 at 18:12
up vote 1 down vote accepted


We have $$1-\sin2x+2\cos2x=0$$

Using Double-Angle Formulas, $$1-\frac{2t}{1+t^2}+2\frac{1-t^2}{1+t^2}=0$$ where $t=\tan x$

Solve the Quadratic Equation in $\tan x$

As $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$

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Thank you , I will try now! – Ivan G. Jan 29 '14 at 18:22
@user123499, yes please revert if you have any confusion – lab bhattacharjee Jan 29 '14 at 18:25

Let $\displaystyle-2=r\cos\phi,1=r\sin\phi$ where $r>0$ so $\displaystyle\frac\pi2<\phi<\pi$ and $\displaystyle\tan\phi=-\frac12$

So we have $$r\cos(2x-\phi)=r\sin\phi$$

$$\implies \cos(2x-\phi)=\cos\left(\frac\pi2-\phi\right)$$

$$\implies 2x-\phi=2n\pi\pm\left(\frac\pi2-\phi\right)$$ where $n$ is any integer

Considering '+' sign, $\displaystyle2x-\phi=2n\pi+\left(\frac\pi2-\phi\right)\implies 2x=2n\pi+\frac\pi2$

Considering '-' sign, $\displaystyle2x-\phi=2n\pi-\left(\frac\pi2-\phi\right)=2n\pi-\frac\pi2+\phi\implies 2x=2n\pi+2\phi-\frac\pi2$

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But can you tell me please, why do we have 2x−ϕ=2nπ±(π/2−ϕ) but not simply 2x−ϕ=π/2−ϕ? – Ivan G. Jan 29 '14 at 18:07
@user123499, as $\cos(-A)=\cos A$ and $\cos(2m\pi+y)=\cos y$ – lab bhattacharjee Jan 29 '14 at 18:11
I know this , but I don't understand why do we write 2x−ϕ=2nπ±(π/2−ϕ) with 2πk? – Ivan G. Jan 29 '14 at 18:15
@user123499, as we have to find $x$ somewhere outside $[0,\frac\pi2]$ – lab bhattacharjee Jan 29 '14 at 18:17
Thanks a lot!!! – Ivan G. Jan 29 '14 at 18:24

from your last step 1−sin2x+2cos2x=0 =>(sin^2x+cos^2x)-2sinxcosx+2(cos^x-sin^2x)=0 =>3cos^2x-2sinxcosx-sin^2x=0 =>3cos^2x-3sinxcosx+sinxcosx-sin^2x=0 =>3cosx(cosx-sinx)+sinx(cosx-sinx)=0 =>(cosx-sinx)(3cosx+sinx)=0 =>cosx-sinx=0 or, 3cosx+sinx=0 =>tanx=1, or tanx=-3 =>x=π/4, or x=2π/3

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You square the both sides


By using double angle identity of $\sin$ and dividing all the terms by $\cos^2(x)$


Quadratic equation

$\tan(x)=1\text{ or }-3$

Find all the solutions in the interval


don't forget to check this value in the original equation.

Here when you substitute this value in the equation it equals zero but if the interval is $[0,2\pi]$, one of the solutions is $\frac{\pi}{4}$ but it doesn't work, it works if the right side is $\sin(x)$ not $-\sin(x)$.

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