Solve for 0 <= x < 2pi.
csc^2 (x) = 4
I know one answer is pi/6. Would the others be 5pi/6, 7pi/6, and 11pi/6?
1/sin^2x = 4
so
sin^2 x = 1/4
so sin x = +1/2 or -1/2
four solutions, yes
yes
To solve the equation csc^2(x) = 4 and find all the solutions in the interval 0 <= x < 2pi, we can begin by taking the reciprocal of both sides:
1/(csc^2(x)) = 1/4
Since csc^2(x) is equivalent to (1/sin(x))^2, we can rewrite the equation as follows:
1/(1/sin^2(x)) = 1/4
Next, we can simplify the equation by multiplying both sides by sin^2(x):
sin^2(x) = 1/4
Now, taking the square root of both sides:
sin(x) = sqrt(1/4)
Simplifying further:
sin(x) = 1/2
We know that sin(x) = 1/2 has solutions at x = pi/6 and x = 5pi/6 on the unit circle (considering the positive y-values only).
However, in the interval 0 <= x < 2pi, we need to find all the solutions. Since sin(x) is positive in the first and second quadrants, we can add the period of sine function, 2pi, to each solution to find additional solutions. Adding 2pi to pi/6 and 5pi/6 gives 13pi/6 and 17pi/6, respectively.
Therefore, the solutions in the interval 0 <= x < 2pi for csc^2(x) = 4 are x = pi/6, 5pi/6, 13pi/6, and 17pi/6.