Question

Solve the following equations.
$cos(\frac{\pi }{2}-3x)-sin2x=0.$

Answer 1

$cos(\frac{\pi }{2}-3x)-sin2x=0.$

$sin3x-sin2x=0....\{\because cos(\frac{\pi }{2}-\theta )=sin\theta \}...\left(1\right)$

$\because sin(A+B)=sinA.cosB+sinB.cosA$

$\because sin(A-B)=sinA.cosB-sinB.cosA$

$\therefore sin(A+B)-sin(A-B)=2sinB.cosA...\left(2\right)$

In this cose,

$A+B=3x$ and $A-B=2x...\left(3\right)$

adding both are get

$A+B=3x$

$A-B=2x$

_________

$2A5x$

$\Rightarrow A=\frac{5x}{2}$ and $B=\frac{x}{2}...\left(4\right)$

from $e{q}^n$ (1), (2), (3) & (4)

$sin3x-sin2x=2sin\left(\frac{x}{2}\right)cos\left(\frac{5x}{2}\right)=0$

$\Rightarrow sin\left(\frac{x}{2}\right)=0$ or $cos\left(\frac{5x}{6}\right)=0$

we know, General solution for

$sin\theta =0\Rightarrow \{\theta =n\pi ,nϵZ\}Z$ integer $\{0,\pm 1,\pm 2,\pm 3...\}$

for $sin\left(\frac{x}{2}\right)\Rightarrow \{\frac{x}{2}=n\pi ,nϵZ\}$

$\Rightarrow \{x=2n\pi ,nϵZ\}$

also we know that, General solution for

$cos\theta =0\Rightarrow \{\theta =(2n+1)\frac{\pi }{2},nϵZ\}$

$\therefore$ for $cos\left(\frac{5x}{2}\right)\Rightarrow \{\frac{5x}{2}=(2n+1)\frac{\pi }{2},nϵZ\}$

$\Rightarrow \{x=(2n+1\left)\frac{\pi }{5,nϵZ}\right\}$

our solution is : $x\left\{\right(2n+1)\frac{\pi }{5},2\pi n/nϵZ\}$