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timurjin [86]
3 years ago
15

Solve sin^2θ−4sinθ−5=0 for 0°≤θ≤360°

Mathematics
1 answer:
Alinara [238K]3 years ago
3 0

Answer:

\theta=270^\circ

Step-by-step explanation:

<u>Given Equation and Parameters</u>

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

<u />

<u>Calculation</u>

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

<u>Conclusion</u>

\theta=270^\circ

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