cos x-sin 3x-cos 2x = 0

### 3 Answers | Add Yours

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.` **

thank you for solving this question

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.

`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes