SOLUTION: An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO

Algebra ->  Trigonometry-basics -> SOLUTION: An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO       Log On


   



Question 1036596: An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
2 sin(3θ) + 1 = 0

Found 2 solutions by Cromlix, ikleyn:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
You have not listed limits
so I am working this out
for 0<=θ=<2π
2 sin(3θ) + 1 = 0
2 sin(3θ) = -1
sin(3θ) = -1/2
(3θ) = π/6, 5π/6
Because you have (3θ) you must
add 2π to each of your values twice:-
(3θ) = π/6, 5π/6,13π/6,17π/6
25π/6 and 29π/6.
Now you divide all the values by 3.
π = π/18, 5π/18,13π/18,17π/18
25π/18 and 29π/18.
Hope this helps :-)

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
2%2Asin%283theta%29+%2B+1 = 0  ---> 

sin%283theta%29) = -1%2F2

3theta = 7pi%2F6+%2B+2k%2Api,   k = 0. +/-1, +/-2, . . .  or/and

3theta = 11pi%2F6+%2B+2k%2Api,  k = 0. +/-1, +/-2, . . .

=====>

theta = 7pi%2F18+%2B+%282k%2Api%29%2F3,   k = 0. +/-1, +/-2, . . .  or/and

theta = 11pi%2F18+%2B+%282k%2Api%29%2F3,  k = 0. +/-1, +/-2, . . .