How do you solve \sin ( 2x ) - \cos ( 3x ) = 0?

1 Answer
Oct 20, 2017

x=npi+(-1)^npi/10 and npi+(-1)^n(13pi)/10 where n belongs to Z.

Explanation:

Here, sin2x-cos3x=0

rarr sin2x=cos3x

rarr 2sinx*cosx=4cos^3x-3cosx

rarr 2sinx*cosx=cosx(4cos^2x-3)

rarr 2sinx=4cos^2x-3

rarr 2sinx=4(1-sin^2x)-3

rarr 2sinx=4-4sin^2x-3

rarr 4sin^2x+2sinx-1=0

Comparing it with ax^2+bx+c=0 we have
a=4 b=2 and c=-1

Using quadratic formula

sinx=(-b+-sqrt(b^2-4ac))/(2*a)

=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)

=(-2+-sqrt(4+16))/8

=(-2+-sqrt(5*2^2))/8

=(-2+-2sqrt(5))/8

Taking + sign we get

sinx=(-2+2sqrt(5))/8

sinx=(sqrt(5)-1)/4

sinx=sin(pi/10)

x=npi+(-1)^npi/10

Taking - sign we get

sinx=(-2-2sqrt(5))/8

sinx=-(sqrt(5)+1)/4

sinx=sin((13pi)/10)

x=npi+(-1)^n(13pi)/10

So the required solution is npi+(-1)^npi/10 and npi+(-1)^n(13pi)/10.