Question

​​​​​​​​​​​​​​​​Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth of a degree, as appropriate.3 cos² θ + 2 cos θ - 1 = 0

Accepted Answer

Recognize that the given equation is a quadratic in terms of $$ \cos 	heta $$: $$ 3 \cos^{2} 	heta + 2 \cos 	heta - 1 = 0 . To $$solve it, let $$ x = \cos 	heta $$, so the equation becomes $$ 3x^{2} + 2x - 1 = 0 . $$Use the quadratic formula to solve for $$ x $$: $$ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$, where $$ a = 3 $$, $$ b = 2 $$, and $$ c = -1 . $$Calculate the discriminant $$ \Delta = b^{2} - 4ac $$ and then find the two possible values for $$ x = \cos 	heta . $$For each value of $$ x $$, determine the corresponding angles $$ 	heta  in $$the interval $$ [0^{\circ}, 360^{\circ})  by $$using the inverse cosine function: $$ 	heta = \cos^{-1}(x) . $$Remember that cosine is positive in Quadrants I and IV, and negative in Quadrants II and III, so find all angles that satisfy each $$ x . $$Write the solutions as exact values if possible, or round to the nearest tenth of a degree as appropriate, ensuring all solutions lie within the given interval.

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth of a degree, as appropriate.
3 cos² θ + 2 cos θ - 1 = 0

Verified video answer for a similar problem:

Key Concepts

Quadratic Equations in Trigonometric Functions

Unit Circle and Angle Measurement

Exact Values and Approximation in Trigonometry