4sin^2(x/2) + cos^2 x = 3 find solutions of this equation in interval [0,2π]?
4sin^2(x/2) + cos^2 x = 3 find solutions of this equation in interval [0,2π]?
find the solutions and if you could explain how to do it because i have the answer i just need to know how to do it
and if someone could give me a link to a good website to study how to find the solutions of this equation
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Use the identity 1 - cos(x) = 2 sin^2(x / 2) to re-write the equation:
2[ 1 - cos(x) ] + cos^2(x) = 3
cos^2(x) - 2 cos(x) - 1 = 0
Solve as a quadratic in cos(x):
cos(x) = [ 2 +/- sqrt(2^2 + 4) ] / 2
= 1 +/- sqrt(2)
cos(x) = 1 + sqrt(2) gives no roots as 1 + sqrt(2) > 1.
cos(x) = 1 - sqrt(2) gives:
x = 180 - 65.5 = 114.5 deg = 1.998 rad.
x = 180 + 65.5 = 245.5 deg = 4.285 rad.
Use the identity 1 - cos(x) = 2 sin^2(x / 2) to re-write the equation:
2[ 1 - cos(x) ] + cos^2(x) = 3
cos^2(x) - 2 cos(x) - 1 = 0
Solve as a quadratic in cos(x):
cos(x) = [ 2 +/- sqrt(2^2 + 4) ] / 2
= 1 +/- sqrt(2)
cos(x) = 1 + sqrt(2) gives no roots as 1 + sqrt(2) > 1.
cos(x) = 1 - sqrt(2) gives:
x = 180 - 65.5 = 114.5 deg = 1.998 rad.
x = 180 + 65.5 = 245.5 deg = 4.285 rad.
I wouldn't like to try finding a website with an equation of exactly that type, but you shouldn't have any problem finding one or more sites which deal with:
(a) trigonometric identities,
(b) quadratic equations,
(c) reference angles.