Asked by Anonymous
Solve for exact or to 1d.p of x E [0 ,2pi]
4cos^2x-3
4cos^2x-3
Answers
Answered by
Answering Life's Questions
4(cosx)^2 - 3 = 0
Add 3 to each side
4(cosx)^2 = 3
Divide both sides by 4
(cosx)^2 = 3/4
Take the square root of each side
cosx = sqrt(3/4) or -sqrt(3/4)
cosx = sqrt(3)/2 or -sqrt(3)/2
x = pi/6, 11pi/6 or 5pi/6, 7pi/6
:)
Add 3 to each side
4(cosx)^2 = 3
Divide both sides by 4
(cosx)^2 = 3/4
Take the square root of each side
cosx = sqrt(3/4) or -sqrt(3/4)
cosx = sqrt(3)/2 or -sqrt(3)/2
x = pi/6, 11pi/6 or 5pi/6, 7pi/6
:)
Answered by
Reiny
To "solve" we need an equation.
I will assume you meant:
4cos^2 x - 3 = 0
cos^2 x = 3/4
cosx = ± √3/2 , ----> all 4 quadrants
you should recognize that cos π/6 = √3/2
x = π/6
or
x = π - π/6 = 5π/6
or
x = π + π/6 = 7π/6
or
x = 2π - π/6 = 11π/6
I will assume you meant:
4cos^2 x - 3 = 0
cos^2 x = 3/4
cosx = ± √3/2 , ----> all 4 quadrants
you should recognize that cos π/6 = √3/2
x = π/6
or
x = π - π/6 = 5π/6
or
x = π + π/6 = 7π/6
or
x = 2π - π/6 = 11π/6
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.