Question

The values of $x$ which satisfies the equation
$sinx+cosx-2\sqrt{2}sinx\cdot cosx=0$ is $x=(2m+1)\frac{\pi }{k}$ when n is even $\left(\frac{2m+1}{3}\right)\pi -\frac{\pi }{3k}$ when n is odd. Find the value of $k$.

Answer 1

$sinx+cosx=2\sqrt{2}sinx.cosx$
$\frac{sin\left(x\right)}{\sqrt{2}}+\frac{cos\left(x\right)}{\sqrt{2}}=2sin\left(x\right)cos\left(x\right)$
$sin(x+\frac{\pi }{4})=sin2x$
Or
$sin\left(2x\right)-sin(x+\frac{\pi }{4})=0$
$2cos(\frac{3x}{2}+\frac{\pi }{8})sin(\frac{x}{2}-\frac{\pi }{8})=0$
Hence
$cos(\frac{3x}{2}+\frac{\pi }{8})=0$
Implies
$\frac{3x}{2}+\frac{\pi }{8}=\frac{(2n+1)\pi }{2}$
Or
$3x+\frac{\pi }{4}=(2n+1)\pi$
$3x=(2n+1)\pi -\frac{\pi }{4}$
$x=\frac{(2n+1)\pi }{3}-\frac{\pi }{12}$...(i)
And
$sin(\frac{x}{2}-\frac{\pi }{8})=0$ implies
$\frac{x}{2}-\frac{\pi }{8}=n\pi$
$x-\frac{\pi }{4}=2n\pi$
$x=2n\pi +\frac{\pi }{4}$...(ii)
Hence from i and ii, we get $k=4$.