Asked by alicia otto
Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<t<2pi.
Solve the equation in the interval [0, 2pi]. The answer must be a multiple of pi
2sin(t)cos(t) + sin(t) -2cos(t)-1=0
Find all solutions of the equation 2cos3x=1
Solve the equation in the interval [0, 2pi]. The answer must be a multiple of pi
2sin(t)cos(t) + sin(t) -2cos(t)-1=0
Find all solutions of the equation 2cos3x=1
Answers
Answered by
Reiny
tan(t)=1/tan (t)
tan^2 t = 1
tan t = ± 1
t = π/4, 3π/4, 5π/4, and 7π/4
2sin(t)cos(t) + sin(t) -2cos(t)-1= 0
by grouping:
sint(2cost + 1) - (2cost + 1) = 0
(2cost+1)(sint-1) = 0
cost = -1/2 or sint = 1
I will let you finish that one
2cos3x = 1
cos3x = 1/2
3x = π/3 or 3x = 2π-π/3 = 5π/3
period of cos3x = 2π/3
so x = π/3 + 2kπ/3
x = 2π/3 + 2kπ/3 , where k is an integer
tan^2 t = 1
tan t = ± 1
t = π/4, 3π/4, 5π/4, and 7π/4
2sin(t)cos(t) + sin(t) -2cos(t)-1= 0
by grouping:
sint(2cost + 1) - (2cost + 1) = 0
(2cost+1)(sint-1) = 0
cost = -1/2 or sint = 1
I will let you finish that one
2cos3x = 1
cos3x = 1/2
3x = π/3 or 3x = 2π-π/3 = 5π/3
period of cos3x = 2π/3
so x = π/3 + 2kπ/3
x = 2π/3 + 2kπ/3 , where k is an integer
Answered by
alicia otto
thanks but it helped some but not much. Reiny
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