Question

URGENT TRIGONOMETRY

Select all that apply.

Solve for x, 0 less than or equal to x less than or equal to 2pi.

2sinx - sqrt3 = 0

a. 5pi/3
b. pi/3
c. 2pi/3
d. 4pi/3

Please explain!

Answer 1

To solve the equation 2sinx - sqrt(3) = 0, we can first add sqrt(3) to both sides to get:

2sinx = sqrt(3)

Then, we can divide both sides by 2 to isolate sin(x):

sinx = sqrt(3)/2

Now, we need to find all values of x that satisfy this equation. We can do this by recalling the values of sin(x) for which it equals sqrt(3)/2. These values occur at x = pi/3 and x = 2pi/3, since sin(pi/3) = sqrt(3)/2 and sin(2pi/3) = sqrt(3)/2. We can also note that sin(x) is positive in the first and second quadrants of the unit circle, where x is between 0 and pi.

Since the problem asks for solutions in the interval 0 ≤ x ≤ 2pi, the correct answers are:

b. pi/3

c. 2pi/3

Therefore, the solutions for x are x = pi/3 and x = 2pi/3. Answers a and d, 5pi/3 and 4pi/3 respectively, are not valid solutions since they fall outside the interval 0 ≤ x ≤ 2pi.