Question #b59e0

2 Answers
Sep 6, 2017

arccos(+-sqrt(3/5))/2

Explanation:

Let X= cos(2x) and Y=sin(2x), the condition i equivalent to
11+Y^2/X^2=7/X^2 => 11X^2+Y^2=7 and X^2+Y^2=1
Subtracting the latest 2 condition we get 10X^2=6
So X^2=3/5 => X=+-sqrt(3/5)
cos(2x)=+-sqrt(3/5)

Sep 8, 2017

30^@; 39^@23

Explanation:

11 + (sin^2 2x)/(cos^2 2x) = 7/(cos 2x)
11cos 2x + (sin^2 2x)/(cos 2x) = 7
Multiply both sides by cos 2x
11 cos ^2 2x + sin^2 2x = 7cos 2x
Replace sin^2 2x by (1 - cos^2 2x)
11cos^2 2x + 1 - cos^2 2x = 7cos 2x
Solve this quadratic equation for cos 2x
10 cos^2 2x - 7cos 2x + 1 = 0
D = d^2 = b^2 - 4ac = 49 - 40 = 9 --> d = +- 3
There are 2 real roots:
cos 2x = - b/(2a) +- d/(2a) = 7/20 +- 3/20
cos 2x = 10/20 = 1/2 and cos 2x = 4/20 = 1/5

a. cos 2x = 1/2 --> 2x = +- pi/3,
x = +- pi/6, or x = +- 30^@
b. cos 2x = 1/5 --> 2x = +- 78^@46 -->
x = +- 39^@23
Answer for (0, 180):
x = 30^@ and x = 39^@23