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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 5sin^(2)theta+9sin theta-5=8sin theta-1 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.\newline5sin⁡2θ+9sin⁡θ−5=8sin⁡θ−1 5 \sin ^{2} \theta+9 \sin \theta-5=8 \sin \theta-1 \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.\newline5sin⁡2θ+9sin⁡θ−5=8sin⁡θ−1 5 \sin ^{2} \theta+9 \sin \theta-5=8 \sin \theta-1 \newlineAnswer: θ= \theta=
  1. Simplify equation: First, let's simplify the given equation by moving all terms to one side to set the equation to zero.5sin⁡2θ+9sin⁡θ−5−8sin⁡θ+1=05\sin^{2}\theta + 9\sin \theta - 5 - 8\sin \theta + 1 = 0
  2. Combine like terms: Now, combine like terms to simplify the equation further.\newline5sin⁡2θ+(9sin⁡θ−8sin⁡θ)−(5−1)=05\sin^{2}\theta + (9\sin \theta - 8\sin \theta) - (5 - 1) = 0\newline5sin⁡2θ+sin⁡θ−4=05\sin^{2}\theta + \sin \theta - 4 = 0
  3. Quadratic formula: We now have a quadratic equation in terms of sin⁡θ\sin \theta. Let's solve for sin⁡θ\sin \theta using the quadratic formula, where a=5a = 5, b=1b = 1, and c=−4c = -4.\newlinesin⁡θ=−b±b2−4ac2a\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  4. Substitute values: Substitute the values of aa, bb, and cc into the quadratic formula.sin⁡θ=−1±12−4(5)(−4)2(5)\sin \theta = \frac{{-1 \pm \sqrt{{1^2 - 4(5)(-4)}}}}{2(5)}sin⁡θ=−1±1+8010\sin \theta = \frac{{-1 \pm \sqrt{{1 + 80}}}}{10}sin⁡θ=−1±8110\sin \theta = \frac{{-1 \pm \sqrt{81}}}{10}
  5. Calculate square root: Calculate the square root and simplify the expression.\newlinesin⁡θ=−1±910\sin \theta = \frac{{-1 \pm 9}}{{10}}\newlineThis gives us two possible solutions for sin⁡θ\sin \theta:\newlinesin⁡θ=(9−1)10=810=0.8\sin \theta = \frac{{(9 - 1)}}{{10}} = \frac{{8}}{{10}} = 0.8\newlinesin⁡θ=(−1−9)10=−1010=−1.0\sin \theta = \frac{{(-1 - 9)}}{{10}} = \frac{{-10}}{{10}} = -1.0
  6. Find angles for sin⁡0.8\sin 0.8: Now we need to find the angles θ\theta that correspond to sin⁡θ=0.8\sin \theta = 0.8 and sin⁡θ=−1.0\sin \theta = -1.0 within the range of 00 degrees to 360360 degrees.\newlineFor sin⁡θ=0.8\sin \theta = 0.8, we use the inverse sine function to find the principal value.\newlineθ=arcsin⁡(0.8)\theta = \arcsin(0.8)
  7. Calculate principal value: Calculate the principal value of θ\theta for sin⁡θ=0.8\sin \theta = 0.8.θ≈arcsin⁡(0.8)≈53.1\theta \approx \arcsin(0.8) \approx 53.1 degrees
  8. Find second angle: Since the sine function is positive in the first and second quadrants, we need to find the second angle that also has a sine of 0.80.8. \newlineθ=180 degrees−53.1 degrees=126.9 degrees\theta = 180 \text{ degrees} - 53.1 \text{ degrees} = 126.9 \text{ degrees}
  9. Find angle for sin⁡−1.0\sin -1.0: For sin⁡θ=−1.0\sin \theta = -1.0, we find the angle where the sine function equals −1-1.\newlineθ=arcsin⁡(−1.0)\theta = \arcsin(-1.0)
  10. Calculate value of theta: Calculate the value of theta for sin⁡θ=−1.0\sin \theta = -1.0.θ=arcsin⁡(−1.0)=270\theta = \arcsin(-1.0) = 270 degrees
  11. Final possible angles: We have found all possible angles for the given equation within the range of 00 degrees to 360360 degrees.\newlineθ=53.1\theta = 53.1 degrees, 126.9126.9 degrees, and 270270 degrees

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