Question

How do I solve for the two smallest positive solutions for: sin(2x)cos(6x)-cos(2x)sin(6x) = -0.35 ?

I understand that I am using a sine sum and difference identity (sin(A+B)=sin(A)cos(B)+cos(A)sin(B)) but I have no clue what to do with the negative decimal number at the end of the equation.

This is what I have so far:
`sin(2x)cos(6x)-cos(2x)sin(6x) = -0.35`

Then I take sine and simplify.
`sin(5x-10x) = -0.35`
`sin(-5x) = -0.35`
`-sin(5x) = - 0.35`

Then I solve for `x`.
`5x = theta`
`x = 1/5theta`

Then I have to solve for `theta`.
`sin(theta) = -0.35`
`sin(theta) = ???`

Once I figure out how to fine `theta`, then I'll be able to find the solutions.

Answer 1

`5^@12; 39^@88`

Explanation:

This equation comes from the trig identity:
sin (a - b) = sin a.cos b - sin b.cos a.
In this case:
sin (2x - 6x) = sin 2x.cos 6x - sin 6x.cos 2x
sin (2x - 6x) = sin (-4x) = - sin 4x = -0.35
sin 4x = 0.35
Calculator and unit circle give 2 solutions for 4x:
a. `4x = 20^@49 + k360^@` -->
`x = 5^@12 + k90^@`, and
b. `4x = 180 - (20.49) = 159^@51 + k360^@` -->
`x = 39^@88 + k90^@`
The 2 smallest positive answers are (k = 0):
`x = 5^@12`, and `x = 39^@88`

Answer 2

`x=5.12^@` or `x=39.88^@`

and in radians `x=0.0894` or `x=0.696`

Explanation:

As `sin(2x)cos(6x)-cos(2x)sin(6x) = -0.35`, we have

`cos(2x)sin(6x)-sin(2x)cos(6x)=0.35`

or `sin(6x)cos(2x)-cos(6x)sin(2x)=0.35` (here we have used commutative property)

As `sin(A-B)=sinAcosB-cosAsinB`, we can write the above as

`sin(6x-2x)=0.35=sin20.49^@`

(We have used scientific calculator to find your `theta` here.)

and hence either `4x=20.49^@` i.e. `x=5.12^@`

or `4x=180^@-20.49^@=159.51^@` i.e. `x=39.88^@`

If you need to find in radians `sin(0.35757)=0.35`

and then `4x=0.35757` or `x=0.0894`

we can also have `4x=pi-0.35757=2.78402` and `x=0.696`

If scientific calculator is not available, one can use tables too.