Hello. I am new to this forum and need some homework help. I have been struggling with trig identities, and can't figure out how to solve sin(3x)+cos(2x)=0. I've made it this far (as x is the only variable, it is sometimes left out):
Solve sin(3x)+cos(2x)=0
(sinxcos2x+cosxsin2x)+cos2x=0
(sinxcos2x+cos(2sincos))+cos2x=0
(sincos2x+2sincos²)+cos2x=0
(sin(2cos²-1)+2sincos²)+cos2x=0
(2sincos²-sin+2sincos²)+cos2x=0
(4sincos²-sin)+cos2x=0
(sin(4cos²-1))+cos2x=0
(sin(2cos+1)(2cos-1))+cos2x=0
From there I could solve the first term alone equaling zero but it's the first term and the second term. I don't know which identity to substitute for cos(2x) as none of them appear to help me get anywhere. If anybody knows of anything I'd greatly appreciate some help.
Solve sin(3x)+cos(2x)=0
(sinxcos2x+cosxsin2x)+cos2x=0
(sinxcos2x+cos(2sincos))+cos2x=0
(sincos2x+2sincos²)+cos2x=0
(sin(2cos²-1)+2sincos²)+cos2x=0
(2sincos²-sin+2sincos²)+cos2x=0
(4sincos²-sin)+cos2x=0
(sin(4cos²-1))+cos2x=0
(sin(2cos+1)(2cos-1))+cos2x=0
From there I could solve the first term alone equaling zero but it's the first term and the second term. I don't know which identity to substitute for cos(2x) as none of them appear to help me get anywhere. If anybody knows of anything I'd greatly appreciate some help.