Hi,

Question

I really need help in understanding how to solve trig equations. How do you solve this equation?

Solve the equation on the interval 0 < or equal to x < or equal to 2pi.

sinx = (square root of 3)/2

I appreciate your help.

Thank you!

Answer 1

you know
(a) sin π/3 = √3/2

now, in the unit circle, sin x = y/r, where r is the radius (hypotenuse)

Since r is always positive, sin x is positive when y is positive, which is in QI and QII

Since sin (π-x) = sin x, sin x = √3/2 when

x = π/3 or (π - π/3) = 2π/3