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.

2 Answers
Nghi N.
Jun 27, 2018

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

Shwetank Mauria
Jun 27, 2018

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.

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