Question

Find the general solution of sin3x + cos2x =o

Answer 1

cos(2x) = -sin(3x)
= -cos(pi/2 - 3x)
= cos(pi/2 + 3x)
Therefore:

Either
2n pi + 2x = pi/2 + 3x
x = (2n - 1/2)pi

or
2n pi - 2x = pi/2 + 3x
x = (2n - 1/2) pi/5


If u can't understand the solution above, then follow the solution given below:

sin 3x+cos2x=0

cos 2x=-sin3x

sin(90–2x)=sin (-3x)

90–2x=n pi+(-1)^n.(-3x)

Case (1) n is even , let n =2m

pi/2–2x=2m.pi-3x

3x-2x=2m.pi-pi/2

x=pi.(2m-1/2) , Answer.

case (2) n is odd , let n=2m+1

90–2x=(2m+1).pi+3x

pi/2-(2m+1).pi=3x+2x

pi.(1/2–2m-1)=5x

pi.(-2m-1/2)= 5x

x=1/5.[pi.(-2m-1/2)] , Answer