Find all solutions to the equation 4sin(x)cos(x)=2sqrt(2)cos(x)
dividing by 4 cos(x) ... sin(x) = √2 / 2
I will interpret your 4sin(x)cos(x)=2sqrt(2)cos(x) as
4sinx*cosx=2√2*cos(x)
4 sinx cosx - 2√2 cosx = 0
cosx(4sinx - 2√2) = 0
cosx = 0 ----> x = π/2, 3π/2 ( or 90°, 270°)
or
sinx = 2√2/4 = √2/2 ----> x = π/4 or 3π/4 (or 45°, 135°)
To find all solutions to the equation 4sin(x)cos(x) = 2√2cos(x), we can use algebraic manipulations and trigonometric identities. Here's how you can approach it:
Step 1: Simplify both sides of the equation.
- Divide both sides of the equation by 2cos(x) to get:
2sin(x) = √2
- Divide both sides of the equation by 2 to get:
sin(x) = √2 / 2
Step 2: Determine the values of x that satisfy the equation.
- Recall that sin(x) = √2 / 2 represents the solutions where the sine of an angle equals √2 / 2.
- In the unit circle, the values of x that satisfy sin(x) = √2 / 2 are π/4 and 3π/4, since the sine function is positive in the first and second quadrants.
- However, we need to find all solutions, so we can use the periodicity of the sine function to determine additional solutions.
- Since sin(x) has a period of 2π, we can add or subtract multiples of 2π to the known solutions obtained above to find more solutions.
- Adding 2π to π/4, we get 9π/4 as another solution.
- Similarly, adding 2π to 3π/4, we get 11π/4 as another solution.
Therefore, the solutions to the equation 4sin(x)cos(x) = 2√2cos(x) are x = π/4, 3π/4, 9π/4, and 11π/4.