Asked by Harry
How would you solve this problem? Please include steps for me!
Solve for θ in the interval 0 ≤ θ ≤ 2π: cos(θ )sin(θ ) − 1/ 2cos(θ ) = 0
Solve for θ in the interval 0 ≤ θ ≤ 2π: cos(θ )sin(θ ) − 1/ 2cos(θ ) = 0
Answers
Answered by
Anonymous
If you mean
cos(θ ) sin(θ ) − (1/ 2) cos(θ ) = 0
that is
cos(θ ) [ sin(θ ) - 0.5 ]
this is true if
cos(θ ) = 0
or if
sin(θ ) = 0.5
cos is zero at π/2 and at 3π/2
sin is zero at 0 and at π ( and at 2π which is zero again of course)
cos(θ ) sin(θ ) − (1/ 2) cos(θ ) = 0
that is
cos(θ ) [ sin(θ ) - 0.5 ]
this is true if
cos(θ ) = 0
or if
sin(θ ) = 0.5
cos is zero at π/2 and at 3π/2
sin is zero at 0 and at π ( and at 2π which is zero again of course)
Answered by
Anonymous
sorry sin = 1/2 at π/6 and at 5 π/6
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