4cos^2 theta = 3 where 0 degrees < theta< 360
4cos^2 theta = 3
cos^2 theta = 3/4
cos (theta) = sqrt(3) / 2
theta = arccos [ sqrt(3) / 2 ]
theta = π/6 , 11π/6
hope this helps~ XD
oops i forgot the +/- sign:
cos (theta) = +/- sqrt(3) / 2
theta = +/- arccos [ sqrt(3) / 2 ]
theta = π/6 , 11π/6 , 5π/6 , 7π/6
To solve the equation 4cos^2(theta) = 3, we can follow these steps:
Step 1: Divide both sides of the equation by 4 to isolate the term with cosine squared:
cos^2(theta) = 3/4
Step 2: Take the square root of both sides of the equation. Remember that we need to consider both the positive and negative square roots:
cos(theta) = ±√(3/4)
Step 3: Simplify the square root:
cos(theta) = ±√3/2
Step 4: Now we need to determine the values of theta that satisfy this equation. To do this, we can recall the values of cosine for different angles in the unit circle. The cosine function represents the x-coordinate of a point on the unit circle.
In the unit circle, the cosine values are positive in the first and fourth quadrants. In the first quadrant (0 to 90 degrees), the cosine is positive. In the fourth quadrant (270 to 360 degrees), the cosine is also positive. Thus, we can say that:
cos(theta) = √3/2, for 0 < theta < 90 degrees
cos(theta) = -√3/2, for 270 < theta < 360 degrees
These are the solutions for theta within the given range of 0 to 360 degrees where the equation 4cos^2(theta) = 3 holds true.