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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -6cos^(2)theta-10 cos theta-1=-3cos theta Answer: theta=

Find all angles, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−6cos⁡2θ−10cos⁡θ−1=−3cos⁡θ -6 \cos ^{2} \theta-10 \cos \theta-1=-3 \cos \theta \newlineAnswer: θ= \theta=

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Q. Find all angles, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−6cos⁡2θ−10cos⁡θ−1=−3cos⁡θ -6 \cos ^{2} \theta-10 \cos \theta-1=-3 \cos \theta \newlineAnswer: θ= \theta=
  1. Simplify the Equation: First, we need to simplify the given equation by moving all terms to one side to form a quadratic equation in terms of cos⁡(θ)\cos(\theta).−6cos⁡2(θ)−10cos⁡(θ)−1=−3cos⁡(θ)-6\cos^{2}(\theta) - 10 \cos(\theta) - 1 = -3\cos(\theta) Add 3cos⁡(θ)3\cos(\theta) to both sides to get:−6cos⁡2(θ)−10cos⁡(θ)+3cos⁡(θ)−1=0-6\cos^{2}(\theta) - 10 \cos(\theta) + 3\cos(\theta) - 1 = 0−6cos⁡2(θ)−7cos⁡(θ)−1=0-6\cos^{2}(\theta) - 7 \cos(\theta) - 1 = 0
  2. Solve Quadratic Equation: Next, we need to solve the quadratic equation for cos⁡(θ)\cos(\theta). We can use the quadratic formula to find the solutions for cos⁡(θ)\cos(\theta). The quadratic formula is given by:\newlinecos⁡(θ)=−b±b2−4ac2a\cos(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlinewhere a=−6a = -6, b=−7b = -7, and c=−1c = -1.
  3. Calculate Discriminant: Calculate the discriminant (b2−4ac)(b^2 - 4ac) to determine if there are real solutions for cos⁡(θ)\cos(\theta).\newlineDiscriminant = (−7)2−4(−6)(−1)(-7)^2 - 4(-6)(-1) = 49−24=2549 - 24 = 25\newlineSince the discriminant is positive, there are two real solutions for cos⁡(θ)\cos(\theta).
  4. Apply Quadratic Formula: Now, apply the quadratic formula to find the values of cos⁡(θ)\cos(\theta).cos⁡(θ)=−(−7)±(25)(2⋅−6)\cos(\theta) = \frac{-(-7) \pm \sqrt{(25)}}{(2 \cdot -6)}cos⁡(θ)=(7±5)−12\cos(\theta) = \frac{(7 \pm 5)}{-12}This gives us two possible values for cos⁡(θ)\cos(\theta):cos⁡(θ)=(7+5)−12=12−12=−1\cos(\theta) = \frac{(7 + 5)}{-12} = \frac{12}{-12} = -1cos⁡(θ)=(7−5)−12=2−12=−16\cos(\theta) = \frac{(7 - 5)}{-12} = \frac{2}{-12} = -\frac{1}{6}
  5. Find Angle for cos⁡(θ)=−1\cos(\theta) = -1: We now need to find the angles θ\theta that correspond to the cos⁡(θ)\cos(\theta) values we found. First, let's find the angles for cos⁡(θ)=−1\cos(\theta) = -1.\newlinecos⁡(θ)=−1\cos(\theta) = -1 occurs at θ=180\theta = 180 degrees.
  6. Find Angle for cos⁡(θ)=−16\cos(\theta) = -\frac{1}{6}: Next, we find the angles for cos⁡(θ)=−16\cos(\theta) = -\frac{1}{6}. We will use a calculator to find the angles to the nearest tenth of a degree. Using the inverse cosine function, we find: θ=cos⁡−1(−16)\theta = \cos^{-1}(-\frac{1}{6}) This will give us two angles, one in the second quadrant and one in the third quadrant, since cosine is negative in both of these quadrants.
  7. Calculate Angles Using Calculator: Calculate the angles using a calculator. \newlineθ≈cos⁡−1(−16)≈99.5\theta \approx \cos^{-1}(-\frac{1}{6}) \approx 99.5 degrees (second quadrant)\newlineθ≈180+(180−99.5)≈260.5\theta \approx 180 + (180 - 99.5) \approx 260.5 degrees (third quadrant)
  8. Finalize the Angles: We have found all the angles that satisfy the given equation within the range 00 to 360360 degrees.\newlineThe angles are approximately 180180 degrees, 99.599.5 degrees, and 260.5260.5 degrees.

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