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Solve sin^2θ−4sinθ−5=0 for 0°≤θ≤360°

Solve sin^2θ−4sinθ−5=0 for 0°≤θ≤360°-example-1

1 Answer

8 votes

Answer:


\theta=270^\circ

Explanation:

Given Equation and Parameters


sin^2\theta-4sin\theta-5=0,\: 0^\circ\leq\theta\leq 360^\circ

Calculation

Let
u=sin\theta, thus:


u^2-4u-5=0\\\\(u+1)(u-5)=0\\\\u=-1,\: u=5


sin\theta=-1\\\\\theta=270^\circ


sin\theta=5\\\\\theta=unde fined

Conclusion


\theta=270^\circ

User Laurencevs
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