SOLUTION: I need to solve for X over interval {0, 2pi)
4cos2x=sqrt3
I tried dividing out 4 but then am stuck with cos2x(which is an identity) = sqrt3/4.
Then im left with 2cos^2x - 1
Algebra ->
Trigonometry-basics
-> SOLUTION: I need to solve for X over interval {0, 2pi)
4cos2x=sqrt3
I tried dividing out 4 but then am stuck with cos2x(which is an identity) = sqrt3/4.
Then im left with 2cos^2x - 1
Log On
Question 293272: I need to solve for X over interval {0, 2pi)
4cos2x=sqrt3
I tried dividing out 4 but then am stuck with cos2x(which is an identity) = sqrt3/4.
Then im left with 2cos^2x - 1 = sqrt3/4.
Should I turn the cos2x into its identity before dividing out 4?
If anyone can help me get on track that would be great! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! I need to solve for X over interval {0, 2pi)
4cos2x=sqrt3
---
cos(2x) = (1/4)sqrt(3)
---
Take the arcos of each side to get:
2x = arcos[(sqrt(3)/4]
---
2x = 64.34 degrees
x = 32.17 degrees
======================
Cheers,
Stan H.
(