Solve the trigonometric equation for the interval 0<x<2pi

Question

4 sin x cos x - 2√3 sin x - 2√2 cos x +√6=0

 

interval 0≤x<2pi

Answer 1

Hi Notroe,

 

Whenever I see a problem like this (asked to solve for x and four terms on the left side and 0 on the right) I get suspicious that factoring by grouping might lead to something.

 

Try:

2sinx(2cosx - √3) - √2(2cosx - √3) = 0   (notice that I had to realize the √6 = √2√3)

 

(2sinx - √2) (2cosx - √3) = 0

 

Now set each factor equal to zero and solve for each x in the target interval.

 

The following graph will help identify the solutions:

 

https://www.desmos.com/calculator/yorzk7ms2f

 

 

Answer 2

4 sin x cos x - 2*sqrt(3) sinx - 2*sqrt(2) cos x + sqrt(6) = 0

Let A = sin x and B = cos x

Then cos x = sqrt( 1 - (sinx)^2 ) =
B = sqrt( 1 - A^2) <--- call this equation ALPHA


The equation becomes:


4 AB - 2sqrt(3)A - 2sqrt(2) B + sqrt(6) = 0

4AB - 2sqrt(2) B = 2sqrt(3)A - sqrt(6) <---- moves terms containing B to left side; everything else to the right

B [ 4A - 2sqrt(2)] = 2sqrt(3)A - sqrt(6) <--- factors out B on left side

B = [ 2sqrt(3)A - sqrt(6) ]/[ 4A - 2sqrt(2)] <--- solves for B

sqrt(1 - A^2) = [ 2sqrt(3)A - sqrt(6) ]/[[ 4A - 2sqrt(2)] <--- substitutes the identity, equation ALPHA, from above

1 - A^2 = [ 12A^2 - 2 * 2sqrt(3)sqrt(6)A + 6 ]/ [16A^2 - 2*8*sqrt(2)*A + 8 ] <--- squares both sides

1 - A^2 = [ 12A^2 - 12*Sqrt(2)*A + 6]/ [16A^2 - 16sqrt(2)*A + 8 ] <--- simplifies the radicals

1-A^2 = (3/4)[ 2A^2 - 2*Sqrt(2)*A + 1]/[2A^2 - 2*sqrt(2)*A + 1] <--- factors out 6 from top, 8 from bottom

1 - A^2 = (3/4) <--- [2A^2 - 2*Sqrt(2)*A + 1] cancels out

1/4 = A^2 <--- A^2 and 3/4 switch places and change sides

A = +or- 1/2 <--- square root of both sides

sin x = +or- 1/2 <--- changes A back to sin x

so x is some "flavor" of 30 degrees since sin 30 = 1/2

The potential candidates for the angles are 30, 150, 210, and 330

Check:
for x=30 sin30=1/2 and cos30 = sqrt(3)/2 <--- 1st quadrant

4(1/2)(sqrt(3)/2) - 2*sqrt(3)*1/2 - 2(sqrt(2))(sqrt(3)/2) + sqrt(6) =
sqrt(3) - sqrt(3) - sqrt(2)*sqrt(3) + sqrt(6) =
-sqrt(6) + sqrt(6) = 0 <--- YES it works
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for x = 150, sin 150 = 1/2 and cos 150 = -sqrt(3)/2 <--- 2nd quadrant

4(1/2)(-sqrt(3)/2) - 2*Sqrt(3)*1/2 - 2*sqrt(2) ( -sqrt(3)/2) + sqrt(6) =
-sqrt(3) - sqrt(3) + sqrt(6) + sqrt(6) <--- No it fails
---------------------------------------------------------------
for x = 210, sin 210 = -1/2 and cos 210 = -sqrt(3)/2 <-- 3rd quadrant

4(-1/2)(-sqrt(3)/2) - 2*Sqrt(3)(-1/2) - 2*sqrt(2)(-sqrt(3)/2)+ sqrt(6) =

sqrt(3) + sqrt(3) + sqrt(6) + sqrt(6) <--- no it fails
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For x = 330, sin330 = -1/2 and cos 330 = sqrt(3)/2

4(-1/2)(sqrt(3)/2) - 2*sqrt(3)(-1/2) - 2*sqrt(2)(sqrt(3)/2) + sqrt(6) =
-sqrt(3) + sqrt(3) - sqrt(6) + sqrt(6) = 0 <--- yes it works!
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So it is the 30 degree angles off of the positive x axis
30 and 330 or pi/6 and 11*pi/6