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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 8sin^(2)theta-11 sin theta-6=-9sin theta-3 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline8sin⁡2θ−11sin⁡θ−6=−9sin⁡θ−3 8 \sin ^{2} \theta-11 \sin \theta-6=-9 \sin \theta-3 \newlineAnswer: θ= \theta=

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Csc, sec, and cot of special anglesright chevron icon

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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.\newline8sin⁡2θ−11sin⁡θ−6=−9sin⁡θ−3 8 \sin ^{2} \theta-11 \sin \theta-6=-9 \sin \theta-3 \newlineAnswer: θ= \theta=
  1. Simplify the equation: First, we need to simplify the given equation by moving all terms to one side to set the equation to zero.\newline8sin⁡2θ−11sin⁡θ−6=−9sin⁡θ−38\sin^{2}\theta - 11\sin \theta - 6 = -9\sin \theta - 3\newlineAdding 9sin⁡θ9\sin \theta and 33 to both sides, we get:\newline8sin⁡2θ−11sin⁡θ+9sin⁡θ−6+3=08\sin^{2}\theta - 11\sin \theta + 9\sin \theta - 6 + 3 = 0\newline8sin⁡2θ−2sin⁡θ−3=08\sin^{2}\theta - 2\sin \theta - 3 = 0
  2. Factor the quadratic equation: Next, we factor the quadratic equation in terms of sin⁡θ\sin \theta. We look for two numbers that multiply to (8)(−3)=−24(8)(-3) = -24 and add up to −2-2. The numbers −6-6 and +4+4 satisfy these conditions. So we can write the equation as: (8sin⁡2θ+4sin⁡θ)−(6sin⁡θ+3)=0(8\sin^{2}\theta + 4\sin \theta) - (6\sin \theta + 3) = 0
  3. Factor by grouping: Now, we factor by grouping.\newlineFirst group: 8sin⁡2θ+4sin⁡θ=4sin⁡θ(2sin⁡θ+1)8\sin^{2}\theta + 4\sin \theta = 4\sin \theta(2\sin \theta + 1)\newlineSecond group: −6sin⁡θ−3=−3(2sin⁡θ+1)-6\sin \theta - 3 = -3(2\sin \theta + 1)\newlineThe equation becomes:\newline4sin⁡θ(2sin⁡θ+1)−3(2sin⁡θ+1)=04\sin \theta(2\sin \theta + 1) - 3(2\sin \theta + 1) = 0
  4. Factor out common factor: We can now factor out the common factor (2sin⁡θ+1)(2\sin \theta + 1):(2sin⁡θ+1)(4sin⁡θ−3)=0(2\sin \theta + 1)(4\sin \theta - 3) = 0
  5. Set factors equal to zero: We set each factor equal to zero and solve for sin⁡θ\sin \theta:2sin⁡θ+1=02\sin \theta + 1 = 0 or 4sin⁡θ−3=04\sin \theta - 3 = 0 For the first equation: 2sin⁡θ=−12\sin \theta = -1 => sin⁡θ=−12\sin \theta = -\frac{1}{2} For the second equation: 4sin⁡θ=34\sin \theta = 3 => sin⁡θ=34\sin \theta = \frac{3}{4}
  6. Find angles for sin⁡θ=−12\sin \theta = -\frac{1}{2}: We find the angles for sin⁡θ=−12\sin \theta = -\frac{1}{2}. Since the sine function is negative in the third and fourth quadrants, we look for angles in those quadrants. The reference angle whose sine is 12\frac{1}{2} is 3030 degrees. Therefore, the angles are 180+30=210180 + 30 = 210 degrees and 360−30=330360 - 30 = 330 degrees.
  7. Find angles for sin⁡θ=34\sin \theta = \frac{3}{4}: We find the angles for sin⁡θ=34\sin \theta = \frac{3}{4}. This is not a standard value for the sine function, so we will use a calculator to find the angle. Using the inverse sine function, we find: θ=sin⁡−1(34)\theta = \sin^{-1}(\frac{3}{4}) θ≈48.6\theta \approx 48.6 degrees (to the nearest tenth) Since the sine function is positive in the first and second quadrants, the other angle is 180−48.6=131.4180 - 48.6 = 131.4 degrees.
  8. Final angles: We now have all the angles that satisfy the equation: θ≈48.6\theta \approx 48.6 degrees, 131.4131.4 degrees, 210210 degrees, and 330330 degrees.

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