Solve −3𝑠𝑖𝑛(4𝑥) − 2 cos(4𝑥) = 1 for the first two positive solutions.

Question

Solve −3𝑠𝑖𝑛(4𝑥) − 2 cos(4𝑥) = 1 for the first two positive solutions.

a. Rewrite −3𝑠𝑖𝑛(4𝑥) − 2 cos(4𝑥) as a single sine function.

b. What is the maximum and minimum value of the single sine function found in part a?

c. Use your rewrote expression from part a to solve for the first two positive solutions.

Answer 1

-3siny -2cosy = 1 (let y=4x)

sin^2y +cos^2y = 1

cos^2y = 1- sin^2y

-3siny -2(sqr(1-sin^2y) =1

-2sqr(1-sin^2y) = 1+3siny square both sides (but you may introduce a 2nd extraneous solution not in the original problem when you multiply by a variable)

4(1-sin^2y) = 1 + 6siny + 9sin^2y

4-4sin^2y = 1 + 6siny + 9siny^2y (let z = siny)

4-4z^2 = 1+ 6z +9z^2

13z^2 + 6z -3 = 0

use the quadratic formula to solve for z

z-.302, -.764 (looks promising as z has to be between -1 and +1; the absolute value of the sine of any angle is never greater than one)

y = the angle whose sine = z

y= 17.58, -49.82

x=y/4 = 4.39,

x=-12.45 degrees

4.39 is the extraneous, false, solution.

check by plugging into the original equation

Answer 2

sin/cos=tan

(-3sin(4x))/(-2cos(4x))-(-2cos(4x))/(-2cos(4x))=1

3/2*tan(4x) - 1 =1

" " =2

" =4/3

4x =53.13°

x =13.2825.........° and 193.2825.........°

Answer 3

a) This will be rather long and ugly, but bear with me here:

cos (4x) = cos (2 *2x)

cos(4x) = cos²(2x) - sin²(2x)

cos(4x) = (cos (2x))² - sin(2x)²

cos (4x) = (1 - 2 sin²x)² - (2 sin(x)cos(x))²

cos (4x) = 1 - 4 sin ²x + 4 sin⁴(x) - (4 sin² x cos²x)

cos (4x) = 1 - 4 sin²x + 4 sin⁴(x) - (4 sin²x (1 - sin²x)

cos (4x) = 1 - 4 sin²x + 4 sin⁴(x) - 4 sin²x + 4 sin⁴x

cos (4x) = 1 - 8 sin²x + 8 sin ⁴(x)

Your original function would then be: -3 sin(4x) -2( 1 - 8 sin²x + 8 sin⁴(x))

b)Max and min for the function would be 2.11 and 2.89

c) -3 sin(4x) - 2 + 16 sin²x - 16 sin⁴x = 1

-3 sin(4x) + 16 sin²x - 16 sin⁴x = -1

-3 sin (4x) + 16 sin²x - 16 sin⁴x = -1

-3 (sin(2*2x)) + 16 sin²x - 16 sin⁴x = - 1

-3 (2 sin(2x) cos(2x)) + 16 sin²x - 16 sin⁴x = -1

-3 (2 (2 sin(x) cos(x) (1 - 2 sin²x) + 16 sin²x - 16 sin⁴x = -1

-3 (4 sin(x) cos (x)(1 - 2 sin²x) + 16 sin²x - 16 sin⁴x = -1

-3 (4 sin (x) cos (x) - 8 sin²x cos (x)) + 16 sin²x - 16 sin⁴x = -1

-12 sin(x) √ 1- sin²x - 8 sin²x √ 1 - sin²x + 16 sin²x - 16 sin⁴x = -1

Honestly, I could go on, but it's gonna get ugly. I just graphed and found the first two solutions, 0.709, 1.351