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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -6sin^(2)theta-9=7sin theta-8 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−6sin⁡2θ−9=7sin⁡θ−8 -6 \sin ^{2} \theta-9=7 \sin \theta-8 \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.\newline−6sin⁡2θ−9=7sin⁡θ−8 -6 \sin ^{2} \theta-9=7 \sin \theta-8 \newlineAnswer: θ= \theta=
  1. Rewrite Equation: Rewrite the given equation in standard quadratic form.\newlineThe given equation is −6sin⁡2θ−9=7sin⁡θ−8-6\sin^{2}\theta - 9 = 7\sin \theta - 8. To rewrite it in standard quadratic form, we need to move all terms to one side of the equation.\newline−6sin⁡2θ−7sin⁡θ−9+8=0-6\sin^{2}\theta - 7\sin \theta - 9 + 8 = 0\newline−6sin⁡2θ−7sin⁡θ−1=0-6\sin^{2}\theta - 7\sin \theta - 1 = 0
  2. Solve Quadratic Equation: Solve the quadratic equation for sin⁡θ\sin \theta. We can use the quadratic formula to solve for sin⁡θ\sin \theta. The quadratic formula is given by sin⁡θ=−b±b2−4ac2a\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=−6a = -6, b=−7b = -7, and c=−1c = -1. sin⁡θ=−(−7)±(−7)2−4(−6)(−1)2(−6)\sin \theta = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(-6)(-1)}}{2(-6)} sin⁡θ=7±49−24−12\sin \theta = \frac{7 \pm \sqrt{49 - 24}}{-12} sin⁡θ=7±25−12\sin \theta = \frac{7 \pm \sqrt{25}}{-12} sin⁡θ=7±5−12\sin \theta = \frac{7 \pm 5}{-12}
  3. Find Possible Values: Find the two possible values for sin⁡θ\sin \theta. We have two possible solutions for sin⁡θ\sin \theta: sin⁡θ=7+5−12=12−12=−1\sin \theta = \frac{7 + 5}{-12} = \frac{12}{-12} = -1 sin⁡θ=7−5−12=2−12=−16\sin \theta = \frac{7 - 5}{-12} = \frac{2}{-12} = -\frac{1}{6}
  4. Find Angle for sin⁡=−1\sin=-1: Find the angles that correspond to sin⁡θ=−1\sin \theta = -1. The angle for which sin⁡θ=−1\sin \theta = -1 is 270°270°.
  5. Find Angle for sin⁡=−16\sin=-\frac{1}{6}: Find the angles that correspond to sin⁡θ=−16\sin \theta = -\frac{1}{6}. To find the angles corresponding to sin⁡θ=−16\sin \theta = -\frac{1}{6}, we use the inverse sine function and consider the symmetry of the sine function in the unit circle. θ=arcsin⁡(−16)\theta = \arcsin(-\frac{1}{6}) Since the sine function is negative in the third and fourth quadrants, we need to find the reference angle in the first quadrant and then find the corresponding angles in the third and fourth quadrants. Reference angle = arcsin⁡(16)≈9.6∘\arcsin(\frac{1}{6}) \approx 9.6^\circ Third quadrant angle = 180∘+9.6∘=189.6∘180^\circ + 9.6^\circ = 189.6^\circ Fourth quadrant angle = 360∘−9.6∘=350.4∘360^\circ - 9.6^\circ = 350.4^\circ

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