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
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) (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) (Show Source): You can put this solution on YOUR website! .