Asked by Miu
Solve each equation for 0<_ 0<_ 2pi(3.14)
A) 2cos^20 - sin0 -1=0
A) 2cos^20 - sin0 -1=0
Answers
Answered by
Reiny
2cos^2 Ø - sinØ -1=0
2(1-sin^2 Ø) - sinØ - 1 = 0
-2sin^2 Ø - sinØ +1 = 0
2sin^2 Ø + sinØ - 1 = 0
(2sinØ + 1)(sinØ - 1) = 0
sinØ = -1/2 or sinØ = 1
case1: sinØ = -1/2
Ø must be in quads III or IV
Ø = 210 or Ø = 330°
or
Ø = 7π/6 or 11π/6
case2:
sinØ = 1
Ø = π/2
2(1-sin^2 Ø) - sinØ - 1 = 0
-2sin^2 Ø - sinØ +1 = 0
2sin^2 Ø + sinØ - 1 = 0
(2sinØ + 1)(sinØ - 1) = 0
sinØ = -1/2 or sinØ = 1
case1: sinØ = -1/2
Ø must be in quads III or IV
Ø = 210 or Ø = 330°
or
Ø = 7π/6 or 11π/6
case2:
sinØ = 1
Ø = π/2
Answered by
haile
solve for theta:2cos(2theta-pi)= -2 ,for -2pi less or equal to theta less than 2pi
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