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Trig Interval?
Find all x in the domain (0,2pi) that satisfy
4cos^2(x)-sin^2(2x)+5sin^2(x)=4
I used double angle formula to get
4+cos(2x) - 1-cos(4x) + 5-cos(2x)
----------- ----------- ----------
2 2 2
And I'm not sure if this is right or if it is where to go from here?
The two are supposed to be spread out and divide each term.
I continued working and divided each by two and got
2 + cos(x) - 1/2 - cos(2x) + 5/2 - cos(x) - 4
then the numbers cancel out 2-0.5+2.5-4=0
Leaves me with
cos(x) - cos(2x) - cos(x)
1 Answer
- ?Lv 76 years ago
4cos^2(x)-sin^2(2x)+5sin^2(x)=4
4cos^2(x) - (2sinxcosx)^2 + 5sin^2(x) - 4 = 0
4cos^2(x) - (4sin^2xcos^2x) + 5sin^2(x) - 4 = 0
4cos^2(x)(1- sin^2x) + 5(1-cos^2x) - 4 = 0
4cos^2(x)(cos^2x) + 5 - 5cos^2x - 4 = 0
4cos^4x - 5cos^2x + 1 = 0
(4cos^2x - 1)(cos^2x -1) = 0
cos^2x = 1/4 => cosx = ±1/2
x = π/3, 2π/3, 4π/3, 5π/3
cos^2x = 1 => cosx = ±1
x = π
