1. Solve: 2 cos² x - 3 cos x + 1 = 0 for 0 ≤ x < 2pi.
2. Solve: 2 sin x - 1 = 0 for 0° ≤ x < 360°
3. Solve: sin² x = cos² x for 0° ≤ x < 360°
4. Solve: sin x - 2sin x cos x = 0 for 0 ≤ x < 2p
1. To solve 2 cos² x - 3 cos x + 1 = 0, we can use factoring or the quadratic formula. Let's use factoring in this case:
The equation can be rewritten as (2cosx - 1)(cosx - 1) = 0.
So, either 2cosx - 1 = 0 or cosx - 1 = 0.
For 2cosx - 1 = 0, we have cosx = 1/2.
For cosx - 1 = 0, we have cosx = 1.
Now we need to find the corresponding values of x for these solutions.
For cosx = 1/2, x can be found using the inverse cosine function (cos⁻¹). Therefore, x = cos⁻¹(1/2).
For cosx = 1, x can be found using the inverse cosine function (cos⁻¹). Therefore, x = cos⁻¹(1).
As mentioned earlier, the values of x need to be between 0 and 2π. Therefore, we need to take the principal value (between 0 and 2π) for x in both cases.
For cos⁻¹(1/2), we get x = π/3 and x = 5π/3.
For cos⁻¹(1), we get x = 0.
So, the solutions for the equation 2 cos² x - 3 cos x + 1 = 0 for 0 ≤ x < 2π are x = π/3, 5π/3, and 0.
2. To solve 2 sin x - 1 = 0 for 0° ≤ x < 360°, we isolate the sin x term:
2 sin x = 1.
Dividing both sides by 2, we get:
sin x = 1/2.
Now we need to find the corresponding values of x for this solution.
Using the inverse sine function (sin⁻¹), we have:
x = sin⁻¹(1/2).
The principal value of x between 0° and 360° for sin⁻¹(1/2) is 30°.
So, the solution for the equation 2 sin x - 1 = 0 for 0° ≤ x < 360° is x = 30°.
3. To solve sin² x = cos² x for 0° ≤ x < 360°, we can use the Pythagorean identity: sin² x + cos² x = 1.
Using this identity, we can rewrite the equation as:
1 - cos² x = cos² x.
Simplifying, we get:
1 = 2cos² x.
Dividing both sides by 2, we get:
1/2 = cos² x.
Taking the square root on both sides, we get:
√(1/2) = cos x.
Since cosine is positive in the first and fourth quadrants, the corresponding values of x for this solution are:
x = √(1/2) or x = -√(1/2).
Using a calculator, we can approximate x = 45° and x = 315°.
So, the solutions for the equation sin² x = cos² x for 0° ≤ x < 360° are x = 45° and x = 315°.
4. To solve sin x - 2sin x cos x = 0 for 0 ≤ x < 2π, we can factor out sin x:
sin x(1 - 2cos x) = 0.
So, either sin x = 0 or 1 - 2cos x = 0.
For sin x = 0, we have x = 0 and x = π.
For 1 - 2cos x = 0, we can isolate cos x:
2cos x = 1.
Dividing both sides by 2, we get:
cos x = 1/2.
Using the inverse cosine function (cos⁻¹), we have:
x = cos⁻¹(1/2).
The principal values of x between 0 and 2π for cos⁻¹(1/2) are π/3 and 5π/3.
So, the solutions for the equation sin x - 2sin x cos x = 0 for 0 ≤ x < 2π are x = 0, x = π, π/3, and 5π/3.
1. To solve the equation 2cos²x - 3cosx + 1 = 0 for 0 ≤ x < 2π, we can use the quadratic formula. The equation is in terms of cos(x), so let's use a substitution to simplify it. Let u = cos(x), then the equation becomes 2u² - 3u + 1 = 0. Now we can solve this quadratic equation for u using the quadratic formula:
u = (-b ± sqrt(b² - 4ac)) / (2a)
For our equation, a = 2, b = -3, and c = 1. Substituting these values in, we get:
u = (-(-3) ± sqrt((-3)² - 4(2)(1))) / (2(2))
= (3 ± sqrt(9 - 8)) / 4
= (3 ± sqrt(1)) / 4
= (3 ± 1) / 4
So we have two solutions for u:
u₁ = (3 + 1) / 4 = 1
u₂ = (3 - 1) / 4 = 1/2
Now we need to solve for x. Since u = cos(x), we can use the inverse cosine function:
x₁ = arccos(1)
x₂ = arccos(1/2)
For 0 ≤ x < 2π, the solutions are x = 0 and x = π/3.
2. To solve the equation 2sinx - 1 = 0 for 0° ≤ x < 360°, we isolate the sine term by moving the constant term to the other side:
2sinx = 1
sinx = 1/2
This is a known trigonometric identity, and we can easily identify that x = 30° or x = 150°, since these are the angles where sin(x) = 1/2. However, since we are given the range as 0° ≤ x < 360°, we need to consider all possible solutions within that range.
In the first quadrant, sin(30°) = 1/2. In the second quadrant, sin(150°) = 1/2. However, sine is also positive in the third quadrant at sin(210°) = 1/2, and in the fourth quadrant at sin(330°) = 1/2.
So the solutions for 0° ≤ x < 360° are x = 30°, 150°, 210°, and 330°.
3. To solve the equation sin²x = cos²x for 0° ≤ x < 360°, we can use the Pythagorean identity for trigonometric functions, which states that sin²x + cos²x = 1. By substituting this identity into the equation, we have:
1 - cos²x = cos²x
By rearranging terms, we get:
2cos²x - 1 = 0
Now, this is a quadratic equation in terms of cos(x). We can solve it using the quadratic formula:
cos(x) = (-b ± sqrt(b² - 4ac)) / (2a)
For our equation, a = 2, b = 0, and c = -1. Substituting these values, we get:
cos(x) = (± sqrt(0 - 4(2)(-1))) / (2(2))
= (± sqrt(8)) / 4
= ± sqrt(2) / 2
Since the cosine function has positive and negative values in multiple quadrants, we have solutions for x at the angles where cos(x) = ± sqrt(2) / 2. These angles are 45°, 135°, 225°, and 315°.
4. To solve the equation sinx - 2sinxcosx = 0 for 0 ≤ x < 2π, we can factor out a sinx term:
sinx(1 - 2cosx) = 0
This equation will be true if either sinx = 0 or (1 - 2cosx) = 0.
For sinx = 0, we know that x = 0 and x = π are solutions.
For 1 - 2cosx = 0, we can solve for cosx:
2cosx = 1
cosx = 1/2
This equation is satisfied for x = π/3 and x = 5π/3.
So the solutions for 0 ≤ x < 2π are x = 0, x = π/3, x = 5π/3, and x = π.