SOLUTION: Hi, I am stuck on this and need a little help. Find all values of x such that 3sinx+1=0 where 0 < or equal to x

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Question 1151900: Hi, I am stuck on this and need a little help. Find all values of x such that 3sinx+1=0 where 0 < or equal to x < or equal to 360 degrees
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the equation is 3 * sin(x) + 1 = 0
subtract 1 from both sides of the equation to get:
3 * sin(x) = -1
divide both sides of the equation by 3 to get:
sin(x) = -1/3
sine is negative in the third and fourth quadrant.
sine is positive. in the first and second quadrant.
in the first quadrant, then the equivalent angle would be sin(x) = 1/3.
solve for x to get x = arcsin(1/3) = 19.471220623 degrees.
the equivalent angle in the second quadrant would be equal to 180 minus that = 160.5287794 degrees.
the equivalent angle in the third quadrant would be equal to 180 plus that = 199.1712206 degrees.
the equivalent angle in the fourth quadrant would be equal to 360 minus that = 340.5287794 degrees.
the sine of the angles will be plus or minus 1/3, depending on the quadrant that the angle is in.
the sinc of the angle in quadrant 1 and 2 should be positive.
the sine of the angle in quadrant 3 and 4 should be negative.
you get:
sin(19.471228623) = 1/3
sin(160.5287794) = 1/3
sin(199.1712206) = -1/3
sin(340.5287794) = -1/3
between 0 and 360 degrees, your solution will be 199.171 degrees and 340.529 degrees, rounded to 3 decimal places.
this can be seen graphically as shown below.
the first graph is the equation of y = sin(x).
here, your solution is when y = -1/3.
the second graph is the equation of y = 3 * sin(x) + 1.
here, your solution is when y = 0.
x represents the angle.
y represents the sine of the angle

SOLUTION: Hi, I am stuck on this and need a little help. Find all values of x such that 3sinx+1=0 where 0 < or equal to x < or equal to 360 degrees