Question

Solve the equation please help?

Answer 1

`x=(npi)/5,(2n+1)pi/2` Where `nrarrZ`

Explanation:

Here, `cosx*cos2x*sin3x=(sin2x)/4`

`rarr2*sin3x[2cos2x*cosx]=sin2x`

`rarr2*sin3x[cos(2x+x)+cos(2x-x)]=sin2x`

`rarr2sin3x[cos3x+cosx]=sin2x`

`rarr2sin3x*cos3x+2sin3x*cosx=sin2x`

`rarrsin6x+sin(3x+x)+sin(3x-x)=sin2x`

`rarrsin6x+sin4x=sin2x-sin2x=0`

`rarrsin6x+sin4x=0`

`rarr2sin((6x+4x)/2)*cos((6x-4x)/2)=0`

`rarrsin5x*cosx=0`

Either, `sin5x=0`

`rarr5x=npi`

`rarrx=(npi)/5`

Or, `cosx=0`

`x=(2n+1)pi/2`

Hence, `x=(npi)/5,(2n+1)pi/2` Where `nrarrZ`