Find all angles, 0^(@)

Question

Find all angles,  0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree.  -2cos^(2)theta-3cos theta=1 Answer:  theta=

Bytelearn · Accepted Answer

Rewrite Equation with Variable: Let's first rewrite the equation in a more standard quadratic form by substituting cos(theta) with a variable, let's say 'x'. So, we have -2x^2 - 3x = 1.Set Equation to Zero: To solve for 'x', we need to set the equation to zero. So, we add 1 to both sides of the equation to get -2x^2 - 3x + 1 = 0.Solve Quadratic Equation: Now, we need to solve the quadratic equation -2x^2 - 3x + 1 = 0. We can use the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = -2, b = -3, and c = 1.Calculate Discriminant: First, we calculate the discriminant, which is b^2 - 4ac. Plugging in the values, we get (-3)^2 - 4(-2)(1) = 9 + 8 = 17.Find First Solution: Now, we can find the two solutions for 'x' using the quadratic formula. The first solution is x = (-(-3) + sqrt(17)) / (2 * -2) = (3 + sqrt(17)) / -4.Find Second Solution: The second solution for 'x' is x = (-(-3) - sqrt(17)) / (2 * -2) = (3 - sqrt(17)) / -4.Check Validity of Solutions: We need to check if these solutions for 'x' are within the range of the cosine function, which is between -1 and 1. Let's calculate the numerical values: x1 = (3 + sqrt(17)) / -4 and x2 = (3 - sqrt(17)) / -4.Calculate Angle for x2: Calculating the values, we get x1 ≈ -1.28 (which is not possible since the range of cosine is between -1 and 1) and x2 ≈ 0.28.Find Second Angle: Since x1 is not within the range of cosine, we discard it. Now, we need to find the angles theta for which cos(theta) = x2 ≈ 0.28. We will use the inverse cosine function to find these angles.Final Angles: Using a calculator, we find that cos^(-1)(0.28) gives us the principal value of theta, which is approximately 73.7 degrees.However, since the cosine function is positive in both the first and fourth quadrants, we need to find the second angle by subtracting the principal value from 360 degrees. So, 360 - 73.7 = 286.3 degrees.We now have two angles that satisfy the original equation: theta ≈ 73.7 degrees and theta ≈ 286.3 degrees. These are the angles between 0 and 360 degrees that satisfy the equation -2cos^2(theta) - 3cos(theta) = 1.

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $-2 \cos ^{2} \theta-3 \cos \theta=1$ $\newline$ Answer: $\theta=$

Full solution

More problems from Find the roots of factored polynomials

Related topics

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−2cos⁡2θ−3cos⁡θ=1 -2 \cos ^{2} \theta-3 \cos \theta=1 −2cos2θ−3cosθ=1\newlineAnswer: θ= \theta= θ=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $-2 \cos ^{2} \theta-3 \cos \theta=1$ $\newline$ Answer: $\theta=$