5cos^2x - 10cosx +5 = 0
please show your work or explain
please show your work or explain
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5cos^2(x) - 10cos(x) + 5 = 0
Factor 5.
5(cos^2(x) - 2cos(x) + 1) = 0
Divide both sides by 5.
cos^2(x) - 2cos(x) + 1 = 0
Factor this like you would factor the quadratic u^2 - 2u + 1; it is a perfect square trinomial.
[cos(x) - 1]^2 = 0
Take the square root of both sides.
cos(x) - 1 = 0
Solve.
cos(x) = 1
Assuming solutions between 0 and 2pi,
x = 0
Factor 5.
5(cos^2(x) - 2cos(x) + 1) = 0
Divide both sides by 5.
cos^2(x) - 2cos(x) + 1 = 0
Factor this like you would factor the quadratic u^2 - 2u + 1; it is a perfect square trinomial.
[cos(x) - 1]^2 = 0
Take the square root of both sides.
cos(x) - 1 = 0
Solve.
cos(x) = 1
Assuming solutions between 0 and 2pi,
x = 0
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Divide by 5.
cos^2x - 2cosx + 1 = (cosx - 1)^2 = 0
cosx = 1
x = 2k pi, where k is any integer
cos^2x - 2cosx + 1 = (cosx - 1)^2 = 0
cosx = 1
x = 2k pi, where k is any integer