SOLUTION: for the interval of 0-2(pi) solve the equation {{{cos4xcos2x+sin4xsin2x=sqrt(3)/2
Algebra.Com
Question 808770: for the interval of 0-2(pi) solve the equation cos4xcos2x+sin4xsin2x=sqrt(3)/2
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Given:
(1) cos(4x)cos(2x) + sin(4x)sin(2x) = sqrt(3)/2
Use the difference formula
(2) cos(a-b) = cos(a)cos(b) + sin(a)sin(b) to obtain
(3) cos(4x - 2x) = cos(4x)cos(2x) + sin(4x)sin(2x)
Now equate (1) and (3) to get
(4) cos(4x - 2x) = sqrt(3)/2 or
(5) cos(2x) = sqrt(3)/2
Use your calculator to evaluate
(6) 2x = arccos(sqrt(3)/2) and get
(7) 2x = 30 or
(8) x = 15 degrees
This is also the solution at
(9) x = 15+180 or
(10) x = 195 degrees
Answer: x = {15,195} degrees
RELATED QUESTIONS
for the interval 0 through 2pi solve the equation... (answered by lwsshak3)
Solve the equation over the interval [0 pi, 2 pi) :... (answered by Fombitz)
Solve the following equation for x:
cos x = -1, for the interval 0 < x < 2... (answered by richard1234)
Solve the equation over the interval [0 pi, 2 pi) :... (answered by robertb)
Please help me solve this equation for all the solutions in the interval [0, 2 pi)... (answered by stanbon)
Solve the equation -3 cos t=1 in the interval from 0 < 0< 2 pi symbol. Round to nearest... (answered by lwsshak3)
solve the equation on the interval 0 < or equal to theta < 2 pi . cot 2theta/3 = -root... (answered by lwsshak3)
Solve on the interval from 0 to pi:2cos^2 x-3 sin... (answered by lwsshak3)
Find all solutions of the equation in the interval
[0, 2 pi)
-sin2x+2cosx=0
(answered by ikleyn)