Cosx-cos2x=sin3x
cos x-sin 3x-cos 2x = 0
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Use the following trigonometric transformation:
`cos x - cos 2x = 2 sin (x+2x)/2*sin (2x-x)/2`
`cos x - cos 2x = 2 sin (3x/2)*sin(x/2)`
Write the middle term sin 3x such that
`sin 3x = sin 2*(3x/2)`
`sin 2*(3x/2) = 2 sin (3x/2)*cos (3x/2)`
Write the new equation:
`2 sin (3x/2)*sin(x/2) - 2 sin (3x/2)*cos (3x/2) = 0`
Notice the common factor `2 sin (3x/2):`
`2 sin (3x/2)*(sin (x/2) - cos(3x/2)) = 0`
`2 sin (3x/2) = 0 =gtsin (3x/2)=0 =gt 3x/2 = arcsin 0 + npi`
`3x/2 = npi =gt x = 2npi/3`
`sin (x/2) - cos(3x/2) = 0`
Use complementary angles to write `sin (x/2)` in terms of `cos (x/2).`
`sin (x/2) = cos (pi/2 - x/2)`
`sin (x/2) - cos(3x/2) = 0 lt=gt cos (pi/2 - x/2) - cos(3x/2) = 0`
Use the trigonometric transformation:
`cos (pi/2 - x/2) - cos(3x/2) = -2 sin (pi/2 - x/2 + 3x/2)/2 *sin (pi/2 - x/2- 3x/2)/2`
If `cos (pi/2 - x/2) - cos(3x/2) = 0 =gt -2 sin (pi/2 - x/2 + 3x/2)/2 *sin (pi/2 - x/2- 3x/2)/2 = 0`
`sin (pi/2 + x) = 0 =gt cos x = 0 =gt x = arccos 0 + 2npi`
`x = pi/2 + 2npi` or `x = 3pi/2 + 2npi`
`sin (pi/2 - 2x) = 0 =gt cos 2x = 0 =gt x = pi/4 + npi or x = 3pi/4 + npi`
The soutions to the given equation are `x = 2npi/3 ; x = pi/2 + 2npi ; x = 3pi/2 + 2npi ; x = pi/4 + npi; x = 3pi/4 + npi.`
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If `cos (pi/2 - x/2) - cos(3x/2) = 0 =gt -2 sin (pi/2 - x/2 + 3x/2)/2 *sin (pi/2 - x/2- 3x/2)/2 = 0`
[sin (pi/2 + x) = 0 =gt cos x = 0 =gt x = arccos 0 + 2npi]
I think that's wrong, because /2 was missed.
it should be: sin((pi/2+x)/2)=0 and sin((pi/2-2x)/2)=0
and then answer is : x=2npi-pi/2 and x=-npi+pi
right?
if no, explain why, please.
`
thank you for solving this question
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