Find all angles, 0^(@)

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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=

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Solve Quadratic Equation: Let's first solve the quadratic equation in terms of cos(theta). The equation is cos²(theta) - cos(theta) - 12 = 0. We can treat cos(theta) as a variable, say 'u', and rewrite the equation as u² - u - 12 = 0. Now we can factor this quadratic equation.Factor Quadratic Equation: To factor the quadratic equation u² - 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 cos²(theta) - cos(theta) - 12 = 0.

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 all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinecos⁡2θ−cos⁡θ−12=0 \cos ^{2} \theta-\cos \theta-12=0 cos2θ−cosθ−12=0\newlineAnswer: θ= \theta= θ=

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: $\theta=$