Asked by Anonymous
whatis the possibilities for the square root of 2 times tan theta = 2 sin theta
Answers
Answered by
Reiny
√2tanØ = 2sinØ
√2 sinØ/cosØ - 2sinØ = 0
sinØ(√2/cosØ - 2) = 0
sinØ = 0 or √2/cosØ - 2 = 0
case 1:
sinØ = 0 , Ø = 0, π, 2π
case 2:
√2/cosØ -2=0
√2/cosØ = 2
2cosØ = √2
cosØ = √2/2
Ø = 45° or 315° or π/4, 7π/4
so for 0 ≤ Ø ≤ 2π
Ø = 0 , π/4 , π , 7π/4, 2π
√2 sinØ/cosØ - 2sinØ = 0
sinØ(√2/cosØ - 2) = 0
sinØ = 0 or √2/cosØ - 2 = 0
case 1:
sinØ = 0 , Ø = 0, π, 2π
case 2:
√2/cosØ -2=0
√2/cosØ = 2
2cosØ = √2
cosØ = √2/2
Ø = 45° or 315° or π/4, 7π/4
so for 0 ≤ Ø ≤ 2π
Ø = 0 , π/4 , π , 7π/4, 2π
Answered by
Steve
An obvious solution is θ=0, since tanθ = sinθ = 0
Eliminating that root, we have
√2/cosθ = 2
cosθ = 1/√2
θ = π/4 or -π/4
Add 2kπ for other roots.
Eliminating that root, we have
√2/cosθ = 2
cosθ = 1/√2
θ = π/4 or -π/4
Add 2kπ for other roots.
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