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Find all angles,
0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree.

sin^(2)theta-4sin theta-5=0
Answer:
theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $\sin ^{2} \theta-4 \sin \theta-5=0$ $\newline$ Answer: $\theta=$

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Find the roots of factored polynomials

Full solution

Q. Find all angles,

0^{\circ} \leq \theta<360^{\circ}

, that satisfy the equation below, to the nearest tenth of a degree.

\newline

\sin ^{2} \theta-4 \sin \theta-5=0

\newline

Answer:

\theta=

Rewrite Equation: Let's first rewrite the given equation to identify it as a quadratic equation in terms of $\sin(\theta)$ : $\sin^2(\theta) - 4\sin(\theta) - 5 = 0$ We can treat $\sin(\theta)$ as a variable, say ' $u$ ', and rewrite the equation as: $u^2 - 4u - 5 = 0$
Factor Quadratic: Now, we will factor the quadratic equation: $\newline$ $(u - 5)(u + 1) = 0$ $\newline$ This gives us two possible solutions for $u$ : $\newline$ $u - 5 = 0$ or $u + 1 = 0$
Solve for u: Solving for 'u' from both equations, we get: $\newline$ $u = 5$ or $u = -1$ $\newline$ Since 'u' was a stand-in for $\sin(\theta)$ , we now have: $\newline$ $\sin(\theta) = 5$ or $\sin(\theta) = -1$
Check Validity: We know that the sine function has a range of $[-1, 1]$ , so $\sin(\theta) = 5$ has no solution because it is outside this range. We only need to consider $\sin(\theta) = -1$ .
Find Angle: The angle $\theta$ that satisfies $\sin(\theta) = -1$ within the range of $0$ to $360$ degrees is: $\theta = 270$ degrees
Final Solution: We have found all the angles that satisfy the given equation. Since $\sin(\theta) = 5$ had no solution, the only angle we found is $\theta = 270$ degrees.

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