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 < 2pi.
1. To solve the equation 2 cos² x - 3 cos x + 1 = 0 for 0 ≤ x < 2π, we can apply the quadratic formula. Let's write the equation in standard form first:
2 cos² x - 3 cos x + 1 = 0
Now, let's assign variables:
a = 2
b = -3
c = 1
Now, using the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values:
x = (-(-3) ± √((-3)² - 4(2)(1))) / (2(2))
x = (3 ± √(9 - 8)) / 4
x = (3 ± √1) / 4
Simplifying further:
x1 = (3 + 1) / 4 = 1
x2 = (3 - 1) / 4 = 1/2
So, the solutions for 0 ≤ x < 2π are x = 1 and x = 1/2.
2. To solve the equation 2 sin x - 1 = 0 for 0° ≤ x < 360°, we need to isolate the sin x term.
First, add 1 to both sides of the equation:
2 sin x = 1
Next, divide both sides by 2:
sin x = 1/2
Now, we can find the angle whose sine is equal to 1/2. In the interval 0° ≤ x < 360°, the angles that satisfy this are 30° and 150°.
Therefore, the solutions for 0° ≤ x < 360° are x = 30° and x = 150°.
3. To solve the equation sin² x = cos² x for 0° ≤ x < 360°, we can use the trigonometric identity:
sin² x + cos² x = 1
So, substituting this equation into the given equation:
sin² x = 1 - sin² x
Now, let's solve for sin x by rearranging the terms:
2 sin² x = 1
sin² x = 1/2
Taking the square root of both sides:
sin x = ±√(1/2)
In the interval 0° ≤ x < 360°, the angles whose sine is equal to ±√(1/2) are 45° and 135°.
Therefore, the solutions for 0° ≤ x < 360° are x = 45° and x = 135°.
4. To solve the equation sin x - 2sin x cos x = 0 for 0 ≤ x < 2π, we can factor out sin x:
sin x(1 - 2cos x) = 0
This equation will be true if either sin x = 0 or (1 - 2cos x) = 0.
First, let's solve sin x = 0. In the given interval, the angles whose sine is equal to 0 are 0 and π.
Next, let's solve 1 - 2cos x = 0. Rearranging the terms:
2cos x = 1
cos x = 1/2
In the given interval, the angles whose cosine is equal to 1/2 are π/3 and 5π/3.
Therefore, the solutions for 0 ≤ x < 2π are x = 0, x = π, x = π/3, and x = 5π/3.
1. To solve the equation 2cos²x - 3cosx + 1 = 0, we can use the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b² - 4ac)) / (2a)
In this equation, a = 2, b = -3, and c = 1. Plugging these values into the quadratic formula, we get:
x = (3 ± √((-3)² - 4(2)(1))) / (2(2))
Simplifying further, we have:
x = (3 ± √(9 - 8)) / 4
x = (3 ± √1) / 4
x = (3 ± 1) / 4
So, the solutions for x are:
x₁ = (3 + 1) / 4 = 1
x₂ = (3 - 1) / 4 = 1/2
Since we are given that 0 ≤ x < 2π, the solutions that fall within this range are 1 and 1/2.
2. To solve the equation 2sinx - 1 = 0, we can isolate the sine term by moving the constant to the other side:
2sinx = 1
Next, we divide both sides of the equation by 2 to get:
sinx = 1/2
To find the value of x within the given range, we can use the inverse sine function (sin⁻¹). This function returns the angle whose sine value is equal to the given value. In this case, sin⁻¹(1/2) = 30° or π/6 radians (since 0° ≤ x < 360°).
So, the solution for x is 30° or π/6 radians.
3. To solve the equation sin²x = cos²x, we can use the Pythagorean identity sin²x + cos²x = 1. Rearranging this equation, we have:
sin²x - cos²x = 0
Factoring the equation, we get:
(sin x + cos x)(sin x - cos x) = 0
To find the solutions, we set each factor equal to zero:
sin x + cos x = 0
sin x - cos x = 0
For sin x + cos x = 0, we can rearrange it as cos x = -sin x. Dividing both sides by cos x (since cos x ≠ 0), we have:
1 = -tan x
This means that x = 3π/4 or x = 7π/4. However, these solutions do not fall within the given range of 0° ≤ x < 360°.
For sin x - cos x = 0, we can rearrange it as sin x = cos x. Dividing both sides by sin x (since sin x ≠ 0), we have:
1 = tan x
This means that x = π/4 or x = 5π/4. Both of these solutions fall within the given range.
So, the solutions for x are π/4 and 5π/4.
4. To solve the equation sin x - 2sin x cos x = 0, we factor out sin x:
sin x(1 - 2cos x) = 0
To find the solutions, we set each factor equal to zero:
sin x = 0
1 - 2 cos x = 0
For sin x = 0, the solutions within the given range of 0 ≤ x < 2π are x = 0 and x = π.
For 1 - 2cos x = 0, we can solve it by isolating cos x:
2cos x = 1
cos x = 1/2
Using the inverse cosine function (cos⁻¹), we find that the angles whose cosine value is 1/2 are 60° and 300°. Converted to radians, they are π/3 and 5π/3.
So, the solutions for x are 0, π, π/3, and 5π/3.