Question

cos 3x - cos 2x = sin 3x find the general solution
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Answer 1

idk

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

cos3x - cos2x = sin3x

=> (cos3x) - (sin3x) = cos2x

=> (4cos^3 x - 3cosx) - (3sinx- 4sin^3 x) = cos2x

=> 4cos^3 x + 4sin^3 x - 3cosx -3sinx = cos2x

=> 4(cos^3 x + sin^3 x) - 3( cosx + sinx) = cos2x

=> 4(cosx + sinx)(cos²x - sinxcosx + sin²x) - 3(cosx+ sinx) = cos2x

=> (cosx + sinx) { 4(1-sinxcosx) - 3} = cos2x

=> (cosx + sinx)(4–4sinxcosx -3) = cos2x

=> (cosx + sinx) (1–4sinxcosx) = (cos²x - sin²x)

=> (cosx + sinx)(1–4sinxcosx)= (cosx+sinx)(cosx-sinx)

=> 1–4sinxcosx-cosx+ sinx = 0

=> 1 - 4 sinx √(1-sin² x) - √(1-sin² x) + sinx = 0

=> sinx + 1 = {√(1-sin²x)} ( 4sinx+ 1)

=> (sinx +1) / (4sinx +1) = √(1-sin²x)

=> (sin² x + 1 + 2sinx) = ( 1-sin² x) ( 16sin² x +1+8sinx)

=> sin²x+1+2sinx = 16sin²x - 16sin^4 x +1 - sin²x +8sinx - 8sin^3 x

=> 16sin^4 x + 8sin^3 x -14sin² x - 6 sinx = 0

=> 2sinx ( 8sin^3 x + 4sin²x - 7sinx - 3) = 0

=> either, 2 sinx = 0

=> sinx = 0

=> x = 0° ……………. (1)

Or, 8sin^3 x +4sin²x - 7sinx -3 = 0

=> 8sin^3 x +4 sin²x - 7sinx = 3

Since 3 = 1+1+1 ( sin values range from 1 to -1 )

=> 8sin^3 x = 1 => sin^3 x = 1/8 => sinx = 1/2

=> x = 30°, which is ruled out, as it doesn't satisfy the given equation.

4sin²x = 1 => sinx = 1/2 => x= 30°, ruled out

Now, 7sinx = -1 => sinx = -1/7 , here also x value is ruled out

Ans: x = 0°