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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-5cos theta=1
Answer: 
theta=

Find all angles, 0θ<360 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline6cos2θ5cosθ=1 -6 \cos ^{2} \theta-5 \cos \theta=1 \newlineAnswer: θ= \theta=

Full solution

Q. Find all angles, 0θ<360 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline6cos2θ5cosθ=1 -6 \cos ^{2} \theta-5 \cos \theta=1 \newlineAnswer: θ= \theta=
  1. Rewrite Equation: Let's first rewrite the equation in a more familiar quadratic form by substituting cos(θ)\cos(\theta) with a variable, let's say 'xx'. So the equation becomes:\newline6x25x=1-6x^2 - 5x = 1\newlineNow, we need to solve for 'xx' before we can find the corresponding angles for θ\theta.
  2. Solve Quadratic Equation: To solve the quadratic equation, we first move all terms to one side to set the equation to zero:\newline6x25x1=0-6x^2 - 5x - 1 = 0\newlineNow, we can use the quadratic formula to find the solutions for 'x'. The quadratic formula is x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=6a = -6, b=5b = -5, and c=1c = -1.
  3. Calculate Discriminant: Let's calculate the discriminant (b24ac)(b^2 - 4ac) first:\newlineDiscriminant = (5)24(6)(1)=2524=1(-5)^2 - 4(-6)(-1) = 25 - 24 = 1\newlineSince the discriminant is positive, we will have two real solutions for x'x'.
  4. Find Solutions for x: Now, we can find the two solutions for 'x' using the quadratic formula:\newlinex=(5)±126x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot -6}\newlinex=5±112x = \frac{5 \pm 1}{-12}\newlineThis gives us two solutions for 'x':\newlinex1=5+112=612=0.5x_1 = \frac{5 + 1}{-12} = \frac{6}{-12} = -0.5\newlinex2=5112=412=13x_2 = \frac{5 - 1}{-12} = \frac{4}{-12} = -\frac{1}{3}
  5. Find Angles for x1x_1: Now we need to find the angles θ\theta that correspond to the cosine values of x1x_1 and x2x_2. Since we are looking for angles between 00 and 360360 degrees, we will use the inverse cosine function and also consider the symmetry of the cosine function.\newlineFor x1=0.5x_1 = -0.5, we find the related angle:\newlineθ1=cos1(0.5)\theta_1 = \cos^{-1}(-0.5)
  6. Calculate θ1\theta_1 and θ2\theta_2: Using a calculator, we find:\newlineθ1=cos1(0.5)120\theta_1 = \cos^{-1}(-0.5) \approx 120 degrees\newlineHowever, since cosine is positive in the fourth quadrant as well, we also have another angle:\newlineθ2=360θ1360120=240\theta_2 = 360 - \theta_1 \approx 360 - 120 = 240 degrees
  7. Find Angles for x2x_2: For x2=13x_2 = -\frac{1}{3}, we find the related angle:\newlineθ3=cos1(13)\theta_3 = \cos^{-1}(-\frac{1}{3})\newlineUsing a calculator, we find:\newlineθ3cos1(13)109.5\theta_3 \approx \cos^{-1}(-\frac{1}{3}) \approx 109.5 degrees (to the nearest tenth)\newlineAgain, considering the symmetry of the cosine function, we find another angle:\newlineθ4=360θ3360109.5=250.5\theta_4 = 360 - \theta_3 \approx 360 - 109.5 = 250.5 degrees (to the nearest tenth)

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