Question #b59e0

2 Answers
Alberto P.
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)#

Nghi N
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#

Related questions
Impact of this question
925 views around the world
You can reuse this answer
Creative Commons License