Solve 3sin2x - 1 = 0
Interval: [0, 2 pi]
Solve 6cos^2x + 5cosx - 6 = 0
For the first one, sin2x = 1/3
2x = 0.33984, 2.80176 or 6.62302 radians (and one more less than 4 pi)
x = 0.16992, 1.4088, or 3.3115
(and one more less than 2 pi that I will leave you to figure out)
For the second question, factor into
(2cosx +3)(3cosx -2) = 0
Then set each factor = 0 and solve for cos x.
To solve the equation 3sin(2x) - 1 = 0, we follow these steps:
Step 1: Add 1 to both sides of the equation to isolate the sine term:
3sin(2x) = 1
Step 2: Divide both sides of the equation by 3:
sin(2x) = 1/3
Step 3: To solve for x, we need to find the inverse sine (sin^(-1)) of both sides:
2x = sin^(-1)(1/3)
Step 4: Evaluate the inverse sine using a calculator to find the angle:
2x = 0.34 radians
Step 5: Divide both sides by 2 to solve for x:
x = 0.34 / 2
Therefore, x ≈ 0.17 radians.
Now, let's solve the equation 6cos^2x + 5cosx - 6 = 0:
This is a quadratic equation, so we'll solve it using the quadratic formula.
Step 1: Identify the coefficients of the quadratic equation:
a = 6, b = 5, c = -6
Step 2: Apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Step 3: Substitute the values into the formula and simplify:
x = (-5 ± √(5^2 - 4*6*(-6))) / (2*6)
x = (-5 ± √(25 + 144)) / 12
x = (-5 ± √169) / 12
x = (-5 ± 13) / 12
Step 4: Solve for both solutions:
x₁ = (-5 + 13) / 12 = 8/12 = 2/3
x₂ = (-5 - 13) / 12 = -18/12 = -3/2
Therefore, the solutions are x = 2/3 and x = -3/2.
To solve the equation 3sin2x - 1 = 0, follow these steps:
Step 1: Write the equation in terms of sine function identity.
3sin2x = 1
Step 2: Divide both sides of the equation by 3.
sin2x = 1/3
Step 3: Use the inverse sine function to solve for 2x.
2x = arcsin(1/3)
Step 4: Apply the periodicity of the sine function.
Since the given interval is [0, 2π], we need to find the general solutions for 2x in this interval.
arcsin(1/3) has a reference angle of π/6, so the initial solution is:
2x = π/6
Step 5: Apply the general solution formula of sin(θ) = sin(π - θ).
Thus, the general solution for 2x is:
2x = π/6 + 2πk or 2x = π - (π/6) + 2πk, where k is an integer.
Step 6: Adjust the general solution to find the solutions in the interval [0, 2π].
Since 2x needs to lie within the [0, 2π] interval, we can narrow down the solutions to:
2x = π/6, 11π/6
Step 7: Divide both sides by 2 to solve for x.
x = π/12, 11π/12
Therefore, the solutions to the equation 3sin2x - 1 = 0 in the interval [0, 2π] are x = π/12 and x = 11π/12.
Now, let's move on to solving the equation 6cos^2x + 5cosx - 6 = 0:
Step 1: This quadratic equation is in the form ax^2 + bx + c = 0, where a = 6, b = 5, and c = -6.
Step 2: Apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values, we get:
x = (-(5) ± √((5)^2 - 4(6)(-6))) / (2(6))
Simplifying further, we have:
x = (-5 ± √(25 + 144)) / 12
x = (-5 ± √169) / 12
x = (-5 ± 13) / 12
Step 3: Simplify the expressions to find the two possible values for x.
First solution:
x = (-5 + 13) / 12
x = 8 / 12
x = 2 / 3
Second solution:
x = (-5 - 13) / 12
x = -18 / 12
x = -3 / 2
Therefore, the solutions to the equation 6cos^2x + 5cosx - 6 = 0 are x = 2/3 and x = -3/2.