Solve on the interval $[0,2 \pi)$. Hint: Factor!
$2 \cos ^2 x+\cos x-1=0$
A. There are no solutions in the given interval.
B. $\left\{0, \frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$
C. $\left\{\frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}\right\}$
D. $\left\{\frac{\pi}{3}, \pi, \frac{5 \pi}{3}\right\}$
E. $\{0\}$
F. $\left\{\frac{\pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}\right\}$
Answer (2)
Factor the trigonometric equation 2 cos 2 x + cos x − 1 = 0 by substituting u = cos x to get ( 2 u − 1 ) ( u + 1 ) = 0 .
Solve for u to find u = 2 1 or u = − 1 .
Substitute back to get cos x = 2 1 or cos x = − 1 .
Find the solutions for x in the interval [ 0 , 2 π ) , which are x = 3 π , π , 3 5 π . Thus, the solution set is { 3 π , π , 3 5 π } .
Explanation
Problem Analysis We are asked to solve the trigonometric equation 2 cos 2 x + cos x − 1 = 0 on the interval [ 0 , 2 π ) . The hint suggests factoring, which is a good approach for quadratic-like trigonometric equations.
Substitution Let's factor the given quadratic equation. We can rewrite the equation as a quadratic in terms of cos x . Let u = cos x . Then the equation becomes 2 u 2 + u − 1 = 0 .
Factoring Now, we factor the quadratic equation 2 u 2 + u − 1 = 0 . We look for two numbers that multiply to 2 ( − 1 ) = − 2 and add to 1 . These numbers are 2 and − 1 . So we can write the middle term as 2 u − u . Thus, 2 u 2 + 2 u − u − 1 = 0 . Factoring by grouping, we get 2 u ( u + 1 ) − 1 ( u + 1 ) = 0 , which simplifies to ( 2 u − 1 ) ( u + 1 ) = 0 .
Solving for u Now we solve for u . We have two cases: 2 u − 1 = 0 or u + 1 = 0 .
Case 1: 2 u − 1 = 0 ⟹ 2 u = 1 ⟹ u = 2 1 .
Case 2: u + 1 = 0 ⟹ u = − 1 .
Substituting Back Now we substitute back cos x for u . So we have cos x = 2 1 or cos x = − 1 .
Finding Solutions for x We need to find the values of x in the interval [ 0 , 2 π ) that satisfy these equations.
For cos x = 2 1 , we know that x = 3 π and x = 2 π − 3 π = 3 5 π are the solutions in the interval [ 0 , 2 π ) .
For cos x = − 1 , we know that x = π is the solution in the interval [ 0 , 2 π ) .
Final Solution Set Therefore, the solution set is { 3 π , π , 3 5 π } .
Examples
Trigonometric equations like this are used in physics to model oscillations and waves. For example, the height of a wave can be modeled using a cosine function, and solving such equations helps determine at what times the wave reaches a certain height. This is crucial in fields like acoustics, optics, and electrical engineering.
To solve the equation 2 cos 2 x + cos x − 1 = 0 , we factor it to find the solutions cos x = 2 1 and cos x = − 1 . This gives us the angles x = 3 π , π , 3 5 π in the interval [ 0 , 2 π ) . The correct answer is B: { 3 π , π , 3 5 π } .
;
