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

cos^(2)theta-cos theta-12=0
Answer:
theta=

Find all angles, $0^{\circ} \leq \theta<360^{\circ}$ , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $\cos ^{2} \theta-\cos \theta-12=0$ $\newline$ 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

\cos ^{2} \theta-\cos \theta-12=0

\newline

Answer:

\theta=

Solve Quadratic Equation: Let's first solve the quadratic equation in terms of $\cos(\theta)$ . The equation is $\cos^2(\theta) - \cos(\theta) - 12 = 0$ . We can treat $\cos(\theta)$ as a variable, say 'u', and rewrite the equation as $u^2 - u - 12 = 0$ . Now we can factor this quadratic equation.
Factor Quadratic Equation: To factor the quadratic equation $u^2 - u - 12 = 0$ , we look for two numbers that multiply to $-12$ and add up to $-1$ . These numbers are $-4$ and $3$ . So we can write the equation as $(u - 4)(u + 3) = 0$ .
Identify Valid Solutions: Now we have two possible solutions for $u$ : $u - 4 = 0$ or $u + 3 = 0$ . Solving for $u$ gives us $u = 4$ or $u = -3$ . However, since $u$ represents $\cos(\theta)$ and the range of the cosine function is $[-1, 1]$ , $u = 4$ is not a valid solution. Therefore, we only consider $u = -3$ , but again, this is outside the range of the cosine function, so there are no solutions in terms of cosine.
Conclude No Solutions: Since there are no valid solutions for $\cos(\theta)$ in the range $[-1, 1]$ , we conclude that there are no angles $\theta$ between $0$ and $360$ degrees that satisfy the original equation $\cos^2(\theta) - \cos(\theta) - 12 = 0$ .

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