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Question Number 79969 by Rio Michael last updated on 29/Jan/20
find the general solution for     2sin 3x = sin 2x
findthegeneralsolutionfor2sin3x=sin2x
Answered by behi83417@gmail.com last updated on 29/Jan/20
2(3sinx−4sin^3 x)=2sinxcosx  ⇒sinx(3−4sin^2 x−cosx)=0  ⇒ { ((sinx=0⇒x=kπ ,k∈Z)),((3−4(1−cos^2 x)−cosx=0)) :}  4cos^2 x−cosx−1=0  ⇒cosx=((1±(√(1+16)))/8)=((1±(√(17)))/8)  ⇒ { ((cosx=((1+(√(17)))/8)⇒x=2mπ±cos^(−1) ((1+(√(17)))/8)[m∈Z])),((cosx=((1−(√(17)))/8)⇒x=2nπ±cos^(−1) ((1−(√(17)))/8)[n∈Z])) :}
2(3sinx4sin3x)=2sinxcosxsinx(34sin2xcosx)=0{sinx=0x=kπ,kZ34(1cos2x)cosx=04cos2xcosx1=0cosx=1±1+168=1±178{cosx=1+178x=2mπ±cos11+178[mZ]cosx=1178x=2nπ±cos11178[nZ]
Commented by Rio Michael last updated on 30/Jan/20
thanks
thanks

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