Question

The solution of the equation
$1+(sinx-cosx)sin\frac{\pi }{4}=2co{s}^2\left(\frac{5}{2}x\right)$, which satisfy the condition $sin6x<0$ is $x=t\pi +\frac{k\pi }{16};tϵz$. Find the value of $k$.

Answer 1

Given,

$1+(sinx-cosx)sin\frac{\pi }{4}=2co{s}^2\left(\frac{5}{2}x\right)$

$\Rightarrow sin\left(x\right)\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}.cos\left(x\right)=2co{s}^2\left(\frac{5x}{2}\right)-1$

$\Rightarrow \sin x\cdot \sin \frac{\pi }{4}-\cos x\cos \cdot \frac{\pi }{4}=2co{s}^2\left(\frac{5x}{2}\right)-1$
$-cos(x+\frac{\pi }{4})=cos\left(5x\right)$ ................ $\because \cos (a+b)=\cos a\cdot \cos b-\sin a\cdot sinb$ and [$\because {\cos }^{2}x=\frac{1+\cos 2x}{2}$]

Now
$sin6x<0$

Hence
$-cos(x+\frac{\pi }{4})=cos5x$
Or
$cos(\pi +\frac{\pi }{4}+x)=cos5x$
$cos(\frac{5\pi }{4}+x)=cos5x$
$\frac{5\pi }{4}+x=5x$
$4x=\frac{5\pi }{4}$
$x=\frac{5\pi }{16}$.

Hence $k=5$.