3 years ago

Solve 11sinθ-2=2 for a value of θ when 0≤θ≤π/2.? How do you solve the second problem?

1 answer:

6 0

5cos (2x)+7=cos (2x)+8, 4cos (2x)=1, cos (2x)=1/4, 2x=arccos(1/4), x=(1/2)arccos (1/4). I used x instead of theta, and arccos is cos inverse

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