Solve 2cos2𝑥+cos𝑥−1=0for all solutions in the interval [0,360°)
2cos2x + cosx - 1 = 0
2(2cos^2x-1) + cosx - 1 = 0
4cos^2x + cosx - 3 = 0
(4cosx-3)(cosx+1) = 0
cosx = 3/4 or -1
since cos41.4° = 3/4,
x = 41.4°, 180°, 360-41.4°
To solve the given equation 2cos^2(x) + cos(x) - 1 = 0 in the interval [0, 360°), we can use a substitution to simplify the equation. Let's substitute cos(x) as t:
t = cos(x)
The equation becomes:
2t^2 + t - 1 = 0
Now, we can solve this quadratic equation for t. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 1, and c = -1. Substituting these values into the formula, we get:
t = (-(1) ± √((1)^2 - 4(2)(-1))) / (2(2))
t = (-1 ± √(1 + 8)) / 4
t = (-1 ± √9) / 4
t = (-1 ± 3) / 4
We have two possible values of t:
t1 = (-1 + 3) / 4 = 2 / 4 = 1/2
t2 = (-1 - 3) / 4 = -4 / 4 = -1
Now, we substitute back cos(x) for t:
For t1:
cos(x) = 1/2
For t2:
cos(x) = -1
To find the solutions in the interval [0, 360°), we need to find the corresponding angles for each value of cos(x). We can use the inverse cosine function (also known as arccosine) to find the angles.
1) cos(x) = 1/2:
Using the inverse cosine function, we find two solutions within the interval [0, 360°):
x = arccos(1/2) ≈ 60°
x = 360° - arccos(1/2) ≈ 300°
2) cos(x) = -1:
Using the inverse cosine function, we find one solution within the interval [0, 360°):
x = arccos(-1) ≈ 180°
Therefore, the solutions to the equation 2cos^2(x) + cos(x) - 1 = 0 in the interval [0, 360°) are:
x ≈ 60°, 180°, and 300°.
To solve the equation 2cos(2x) + cos(x) - 1 = 0 in the interval [0, 360°), we can use the following steps:
Step 1: Identify the trigonometric identities or any simplifications that can be made to put the equation in a simpler form.
In this equation, there is no obvious trigonometric identity or simplification that can be made, so we move on to the next step.
Step 2: Use substitution to solve the equation.
Let's substitute u = cos(x) to simplify the equation:
2cos(2x) + cos(x) - 1 = 0
Substituting u = cos(x), we have:
2cos(2x) + u - 1 = 0
Step 3: Solve the equation using algebraic methods.
Let's rewrite the equation in terms of u:
2(2u^2 - 1) + u - 1 = 0
Expanding and simplifying:
4u^2 - 2 + u - 1 = 0
4u^2 + u - 3 = 0
Now, we can factorize the equation:
(4u - 1)(u + 3) = 0
Setting each factor to zero and solving for u:
4u - 1 = 0 → 4u = 1 → u = 1/4
u + 3 = 0 → u = -3
So we have two possible values for u: u = 1/4 and u = -3.
Step 4: Solve for x.
Now, let's solve for x using the values of u we found:
For u = 1/4:
cos(x) = 1/4
Taking the inverse cosine (cos^-1) of both sides:
x = cos^-1(1/4)
Using a calculator to find the principal value of cos^-1(1/4), we get:
x ≈ 75.52°
For u = -3:
cos(x) = -3
This equation has no real solutions since the cosine function can only have values between -1 and 1.
Therefore, the only solution in the interval [0, 360°) is x ≈ 75.52°.