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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+6sin theta=-1 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+6sin⁡θ=−1 8 \sin ^{2} \theta+6 \sin \theta=-1 \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+6sin⁡θ=−1 8 \sin ^{2} \theta+6 \sin \theta=-1 \newlineAnswer: θ= \theta=
  1. Rewrite Equation in Quadratic Form: Rewrite the given equation in a quadratic form by substituting sin⁡(θ)\sin(\theta) with a variable, let's say 'xx'. This gives us the equation 8x2+6x+1=08x^2 + 6x + 1 = 0.
  2. Solve Quadratic Equation: Solve the quadratic equation 8x2+6x+1=08x^2 + 6x + 1 = 0 using the quadratic formula, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=8a = 8, b=6b = 6, and c=1c = 1.
  3. Calculate Discriminant: Calculate the discriminant, which is b2−4ac=62−4(8)(1)=36−32=4b^2 - 4ac = 6^2 - 4(8)(1) = 36 - 32 = 4.
  4. Find Solutions for xx: Since the discriminant is positive, there are two real solutions for xx. Calculate these solutions using the quadratic formula:\newlinex=−6±42⋅8x = \frac{{-6 \pm \sqrt{4}}}{{2 \cdot 8}}\newlinex=−6±216x = \frac{{-6 \pm 2}}{{16}}
  5. Find Angles for sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}: Find the two values of xx:x1=(−6+2)16=−416=−14x_1 = \frac{(-6 + 2)}{16} = \frac{-4}{16} = -\frac{1}{4}x2=(−6−2)16=−816=−12x_2 = \frac{(-6 - 2)}{16} = \frac{-8}{16} = -\frac{1}{2}
  6. Find Angles for sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}: Recall that xx represents sin⁡(θ)\sin(\theta). Since the range of the sine function is [−1,1][-1, 1], both solutions x1=−14x_1 = -\frac{1}{4} and x2=−12x_2 = -\frac{1}{2} are valid for sine values. We need to find the corresponding angles θ\theta for these sine values.
  7. Find Angles for sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}: Recall that xx represents sin⁡(θ)\sin(\theta). Since the range of the sine function is [−1,1][-1, 1], both solutions x1=−14x_1 = -\frac{1}{4} and x2=−12x_2 = -\frac{1}{2} are valid for sine values. We need to find the corresponding angles θ\theta for these sine values.Find the angles corresponding to sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}. Since the sine function is negative in the third and fourth quadrants, we look for angles in those quadrants. Use the inverse sine function to find the reference angle, and then find the actual angles.\newlineθ1=arcsin⁡(−14)≈−14.5\theta_1 = \arcsin(-\frac{1}{4}) \approx -14.5 degrees (reference angle)\newlineThe actual angles are:\newlineθ1=180+14.5≈194.5\theta_1 = 180 + 14.5 \approx 194.5 degrees (third quadrant)\newlinexx00 degrees (fourth quadrant)
  8. Find Angles for sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}: Recall that xx represents sin⁡(θ)\sin(\theta). Since the range of the sine function is [−1,1][-1, 1], both solutions x1=−14x_1 = -\frac{1}{4} and x2=−12x_2 = -\frac{1}{2} are valid for sine values. We need to find the corresponding angles θ\theta for these sine values.Find the angles corresponding to sin⁡(θ)=−14\sin(\theta) = -\frac{1}{4}. Since the sine function is negative in the third and fourth quadrants, we look for angles in those quadrants. Use the inverse sine function to find the reference angle, and then find the actual angles.θ1=arcsin⁡(−14)≈−14.5\theta_1 = \arcsin(-\frac{1}{4}) \approx -14.5 degrees (reference angle)The actual angles are:θ1=180+14.5≈194.5\theta_1 = 180 + 14.5 \approx 194.5 degrees (third quadrant)xx00 degrees (fourth quadrant)Find the angles corresponding to sin⁡(θ)=−12\sin(\theta) = -\frac{1}{2}. Again, since the sine function is negative in the third and fourth quadrants, we look for angles in those quadrants. Use the inverse sine function to find the reference angle, and then find the actual angles.xx22 degrees (reference angle)The actual angles are:xx33 degrees (third quadrant)xx44 degrees (fourth quadrant)

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