# Question 6527d

Mar 1, 2018

$x = \frac{\pi}{4} , \frac{7 \pi}{12} , \frac{11 \pi}{12.}$

#### Explanation:

$\sin 3 x + \cos 3 x = 0$
$\Rightarrow \cos \left(\frac{\pi}{2} - 3 x\right) + \cos 3 x = 0$
$\Rightarrow 2 \cos \left(\frac{\frac{\pi}{2} - 3 x + 3 x}{2}\right) \cos \left(\frac{\frac{\pi}{2} - 3 x - 3 x}{2}\right) = 0$
$\Rightarrow 2 \cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4} - 3 x\right) = 0$
$\Rightarrow \cos \left(\frac{\pi}{4} - 3 x\right) = 0 , w h e r e , 2 \cos \left(\frac{\pi}{4}\right) \ne 0$
$\Rightarrow \cos \left(3 x - \frac{\pi}{4}\right) = 0$, as $\cos \left(- \theta\right) = \cos \theta$
$\Rightarrow 3 x - \frac{\pi}{4} = \left(2 k + 1\right) \frac{\pi}{2} , k \in Z ,$
$\Rightarrow 3 x = \left(2 k + 1\right) \frac{\pi}{2} + \frac{\pi}{4}$
$\textcolor{red}{\Rightarrow x = \left(2 k + 1\right) \frac{\pi}{6} + \frac{\pi}{12} , k \in Z}$,
So, taking $k = 0 , \pm 1. \pm 2 , \pm 3. . .$
k=0rArrx=pi/6+pi/12=color(blue)(pi/4in(0.pi),
$k = 1 \Rightarrow x = \frac{3 \pi}{6} + \frac{\pi}{12} = \textcolor{b l u e}{\frac{7 \pi}{12} \in \left(0. \pi\right)}$
 k=2rArrx=(5pi)/6+pi/12=color(blue)((11pi)/12in(0.pi),
 k=3rArrx=(7pi)/6+pi/12=(15pi)/12!in(0.pi)
Also,$k = - 1 \Rightarrow x = \frac{- \pi}{6} + \frac{\pi}{12} = - \frac{\pi}{12} \notin \left(0. \pi\right)$
Hence, $x = \frac{\pi}{4} , \frac{7 \pi}{12} , \frac{11 \pi}{12}$

Apr 30, 2018

$x = \frac{\pi}{4} , \frac{7 \pi}{12} , \frac{11 \pi}{12}$
We know that,
$\textcolor{b l u e}{\left(I\right) \sin 3 \theta = 3 \sin \theta - 4 {\sin}^{3} \theta}$
$\textcolor{b l u e}{\left(I I\right) \cos 3 \theta = 4 {\cos}^{3} \theta - 3 \cos \theta}$
color(violet)((III)sin2theta=2sinthetacostheta

#### Explanation:

Here,

color(blue)(sin3x+cos3x=0) ,where,color(red)(x in (0,pi)

Using above color(blue)((I)and (II)

color(blue)(3sinx-4sin^3x+4cos^3x-3cosx=0

$\implies 4 {\cos}^{3} x - 4 {\sin}^{3} x - 3 \cos x + 3 \sin x = 0$

$\implies 4 \left({\cos}^{3} x - {\sin}^{3} x\right) - 3 \left(\cos x - \sin x\right) = 0$

$\implies 4 \left(\cos x - \sin x\right) \left({\cos}^{2} x + \cos x \sin x + {\sin}^{2} x\right) - 3 \left(\cos x - \sin x\right) = 0$

$\implies \left(\cos x - \sin x\right) \left[4 \left({\cos}^{2} x + \cos x \sin x + {\sin}^{2} x\right) - 3\right] = 0$

$\implies \left(\cos x - \sin x\right) \left[4 \left(1 + \cos x \sin x\right) - 3\right] = 0$

$\implies \left(\cos x - \sin x\right) \left[4 + 4 \sin x \cos x - 3\right] = 0$

$\implies \left(\cos x - \sin x\right) \left(1 + 2 \cdot 2 \sin x \cos x\right) = 0$

=>(cosx-sinx)(1+2color(violet)(sin2x))=0...tocolor(violet)(Apply(III)

$\implies \cos x - \sin x = 0 \mathmr{and} 1 + 2 \sin 2 x = 0 \implies \sin 2 x = - \frac{1}{2}$

Now, $x \in \left(0 , \pi\right) \implies {I}^{s t} Q u a \mathrm{dr} a n t \mathmr{and} I {I}^{n d} Q u a \mathrm{dr} a n t$

$\left(i\right) \cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 > 0$

i.e.I^(st)Quadrant=>color(red)(x=pi/4

Also, $x \in \left(0 , \pi\right)$

$\implies 2 x \in \left(0 , 2 \pi\right) \implies {I}^{s t} , I {I}^{n d} , I I {I}^{r d} , I {V}^{t h} Q u a \mathrm{dr} a n t$

$\left(i i\right) \sin 2 x = - \frac{1}{2} < 0 \implies I I {I}^{r d} \mathmr{and} I {V}^{t h} Q u a \mathrm{dr} a n t$

$\therefore 2 x = \pi + \frac{\pi}{6} = \frac{7 \pi}{6} \mathmr{and} 2 x = 2 \pi - \frac{\pi}{6} = \frac{11 \pi}{6}$

i.e. color(red)( x=(7pi)/12 or x=(11pi)/12#

Hence, $x = \frac{\pi}{4} , \frac{7 \pi}{12} , \frac{11 \pi}{12}$