Determine the general solution of
4cos^2x+2sinxcosx-1=0
To determine the general solution of the given equation, we can follow these steps:
Step 1: Rearrange the equation
Rearrange the given equation to isolate the trigonometric function. In this case, we will isolate the cosine function:
4cos^2x + 2sinxcosx - 1 = 0
Step 2: Use the trigonometric identity
Use the trigonometric identity cos^2(x) + sin^2(x) = 1 to substitute for sin^2(x):
4(cos^2(x) + sin^2(x)) - 2sinxcosx - 1 = 0
4cos^2(x) + 4sin^2(x) - 2sinxcosx - 1 = 0
4cos^2(x) + (4 - 4cos^2(x)) - 2sinxcosx - 1 = 0
-4cos^2(x) - 2sinxcosx + 3 = 0
Step 3: Factor the equation
Factor the equation to simplify it:
-(2cosx + 1)(2cosx - 3) = 0
Step 4: Set each factor equal to zero and solve
Set each factor equal to zero and solve for x:
2cosx + 1 = 0 or 2cosx - 3 = 0
For 2cosx + 1 = 0:
2cosx = -1
cosx = -1/2
For 2cosx - 3 = 0:
2cosx = 3
cosx = 3/2, which is not a valid solution since the range of cosine function is between -1 and 1.
Step 5: Find the solutions in the given interval
Determine the solutions for cosx = -1/2 in the given interval. The general solution for cosx = -1/2 can be found using the unit circle or the trigonometric values table.
From the unit circle, we can see that cosx = -1/2 has solutions at x = 2π/3 + 2nπ and x = 4π/3 + 2nπ, where n is an integer.
Therefore, the general solution for the given equation is:
x = 2π/3 + 2nπ or x = 4π/3 + 2nπ, where n is an integer.