What are the solutions for 3cos2x=-3 if 0≤ x ≤ 2pi
To solve the equation 3cos2x = -3, we'll follow these steps:
Step 1: Divide both sides of the equation by 3:
cos2x = -1
Step 2: Recall the identity for double angles:
cos2x = 2cos^2(x) - 1
Step 3: Substitute the expression for cos2x into the equation:
2cos^2(x) - 1 = -1
Step 4: Rearrange the equation:
2cos^2(x) = 0
Step 5: Divide both sides by 2:
cos^2(x) = 0
Step 6: Take the square root of both sides:
cos(x) = 0
Step 7: Find the values of x where cos(x) is equal to 0 within the given interval of 0≤ x ≤ 2pi.
The values of x where cos(x) is equal to 0 within the given interval are:
x = π/2 and x = 3π/2
So, the solutions for the equation 3cos2x = -3 within the interval 0≤ x ≤ 2pi are:
x = π/2 and x = 3π/2
To find the solutions for 3cos(2x) = -3, we will apply a step-by-step process:
Step 1: Simplify the equation.
Divide both sides of the equation by 3:
cos(2x) = -1
Step 2: Determine the values of x that satisfy cos(2x) = -1.
The cosine function reaches its minimum values of -1 at certain intervals on the unit circle. Those intervals are when the angle is π radians (180 degrees) plus an even multiple of π radians (360 degrees). In other words, we have:
2x = π + 2nπ, where n is an integer.
Step 3: Solve for x.
Divide both sides by 2 to isolate x:
x = (π + 2nπ)/2
x = π/2 + nπ, where n is an integer.
Step 4: Determine the solutions within the given range.
Since the range is 0 ≤ x ≤ 2π, we need to find the values of x that satisfy this condition.
For n = 0: x = π/2
For n = 1: x = π/2 + π = 3π/2
For n = 2: x = π/2 + 2π = 5π/2
Since 5π/2 is greater than 2π, it falls outside the range. Therefore, the solutions within the given range are x = π/2 and x = 3π/2.
So, the solutions for 3cos(2x) = -3, where 0 ≤ x ≤ 2π, are x = π/2 and x = 3π/2.
That is Trigonometry problem
Trigonometry isn't Calculus
3 cos 2x = - 3
Divide both sides by 3
cos 2x = - 1
2 x = cos⁻¹ ( - 1 )
2 x = arccos ( - 1 )
arccos ( - 1 ) = π
The period of cosine function is 2 π so:
2 x = π + 2 π n
Divide both sides by 2
x = π / 2 + π n
n = 0 , ±1 , ±2 , ±3...
For 0 ≤ x ≤ 2 π
x = π / 2 + π ∙ 0 = π / 2 + 0 = π / 2
and
x = π / 2 + π ∙ 1 = π / 2 + π = π / 2 + 2 π / 2 = 3 π / 2