SOLUTION: Solve the equation for 0 <= θ < 2 pi. What is -6 sin θ = 3 square root of 3.

Algebra.Com
Question 1163783: Solve the equation for 0 <= θ < 2 pi. What is -6 sin θ = 3 square root of 3.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
 0 <= θ < 2 pi. What is -6 sin θ = 3 square root of 3.



Divide both sides by -6





The sine is positive only in QIII and QIV



Edwin
RELATED QUESTIONS

solve the following equation for all angles in [0,2pi) sin(2x-pi/3)= square root... (answered by lwsshak3)
Determine an exact value for the expression {{sin pi/3 - cos 5 pi/6} divided by {1 - tan... (answered by jsmallt9)
Please solve the equation sin*(2*x+(pi/3)) = 1/2 for x is [0,pi] Show... (answered by josmiceli)
Evaluate: sin 2pi cos pi/3 - cos 2pi sin pi/3 P.S. the answer is -square root of 3/2... (answered by jsmallt9)
2 sin x/2 + square root 3 = 0 solve the multiple angle equation (answered by ikleyn)
I have several questions. 1. Find a root of 1/2*x^2 + x + 1 - e^x = 0 2. Find two of (answered by khwang)
solve and state the four exact solutions on the interval [0, 2 pi]: 2 sin (2 x)+ square... (answered by oscargut)
I need help with these questions! Thank you so much! 1. Express sec 0- sin0tan0 in terms (answered by rwm)
what is the solution for square root of 2/3 times square root 5/6 (answered by richwmiller)