how do i solve these problems in the interval [0,2π]?
it says give answer as a multiple of π. do not use decimals. the answer should be a fraction or integer. if there is more than one answer separate them by commas.
(sin(t))^2 = 1/2
| tan(t) |=1
2(cos(t))^2 - cos(t)-1=0
tan(t)=√(3)
tan(t)=1/√(3)
Thank you!!
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Verified answer
In the interval [0, 2π],
(sin(t))^2 = 1/2
sin t = ± 1/sqrt 2
t = π/4, 3π/4, 5π/4, 7π/4
| tan(t) |=1
tan t = ± 1
t = π/4, 3π/4, 5π/4, 7π/4
2(cos(t))^2 - cos(t)-1=0
This is a quadratic equation in cos t.
By the quadratic formula,
cos t = {1 ± sqrt [1 - 4*2*(-1)]}/(2*2)
= (1 ± sqrt 9)/4
= (1 ± 3)/4
= 1 or -1/2
t = 0, 2π, 2π/3, 4π/3
tan(t)=√(3)
t = π/3, 4π/3
tan(t)=1/√(3)
t = π/6, 7π/6