Trigonometry> how to solve cos3x + cos2x = sin3x/2 + si...

how to solve cos3x + cos2x = sin3x/2 + sin x/2 , 0

Mohammed Saleem Mohiuddin , 9 Years ago

Grade 12th pass

4 Answers

Vikas TU

I think you're doing something like the following:

Consider this problem which is similar to yours:

1+sinx-sin2x=cosx+cos2x-cos3x :

1+sinx-sin2x=cosx+cos2x-cos3x
move cosx and cos3x to the left side
1+sinx-sin2x+cos3x-cosx=cos2x
cos3x-cosx=-2sin2x*sinx and cos2x=cos^2x-sin^2x and 1=sin^2x+cos^2x
subtract 1 from both sides
sinx-sin2x-2sin2x*sinx=cos^2x-sin^2x-(sin^2x+cos^2x)=>cos^2x-sin^2x-sin^2x-cos^2x
sinx-sin2x-2sin2x*sinx=-2sin^2x
add -2sin^2x for both sides
2sin^2x+sinx-sin2x-2sin2x*sinx=0
sin2x=2sinxcosx
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sinx*(2sinx+1-2cosx-2sin2x)=0 if sinx=0 x=k*pi
or 2(sinx-cosx-sin2x)+1=0

Last Activity: 9 Years ago

naveen

I think you're doing something like the following:

Consider this problem which is similar to yours:

1+sinx-sin2x=cosx+cos2x-cos3x :

Last Activity: 9 Years ago

raj

I think you're doing something like the following:

Consider this problem which is similar to yours:

1+sinx-sin2x=cosx+cos2x-cos3x :

1+sinx-sin2x=cosx+cos2x-cos3x
move cosx and cos3x to the left side
1+sinx-sin2x+cos3x-cosx=cos2x
cos3x-cosx=-2sin2x*sinx and cos2x=cos^2x-sin^2x and 1=sin^2x+cos^2x
subtract 1 from both sides
sinx-sin2x-2sin2x*sinx=cos^2x-sin^2x-(sin^2x+cos^2x)=>cos^2x-sin^2x-sin^2x-cos^2x
sinx-sin2x-2sin2x*sinx=-2sin^2x
add -2sin^2x for both sides
2sin^2x+sinx-sin2x-2sin2x*sinx=0
sin2x=2sinxcosx
||
hello,
sinx*(2sinx+1-2cosx-2sin2x)=0 if sinx=0 x=k*pi
or 2(sinx-cosx-sin2x)+1=0

Last Activity: 9 Years ago

Man Dhungana

Here, cos2x + cos3x = sin3x/2 + sinx/2.or, 2cos5x/2 . cosx/2 = 2 sinx.cosx/2 [ Since, CosC + CosD = 2Cos(C+D)/2.Cos[(C-D)/2 & SinC + SinD = 2Sin(C+D)/2.Cos(C-D)/2.]or, cos5x/2 . cosx/2 - sinx.cosx/2 = 0.or, cosx/2( cos 5x/2 - sinx) = 0Either, Or, cosx/2 = 0 cos5x/2 - sinx = 0Now solve few steps and find the answer.

Last Activity: 7 Years ago