Asked by sherri
8 Tan^2 theta sin theta + 4 Tan^2 theta = 0
Answers
Answered by
Steve
fcator out 4tan^2θ to get
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = -π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = π/6 + +2πn
for any integer n
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = -π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = π/6 + +2πn
for any integer n
Answered by
Steve
fcator out 4tan^2θ to get
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = 7π/6 or θ = 11π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = 7π/6 + +2πn
θ = 11π/6 + +2πn
for any integer n
4tan^2θ(2sinθ+1) = 0
so, tanθ = 0 or sinθ = -1/2
θ=0 or θ = 7π/6 or θ = 11π/6
You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be
θ = πn
θ = 7π/6 + +2πn
θ = 11π/6 + +2πn
for any integer n
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