SOLUTION: Please help me solve this question.
Solve each of the following for 0 <= theta <= 360 degrees
3sin^2 theta + 2sin theta = 0
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Question 1180791: Please help me solve this question.
Solve each of the following for 0 <= theta <= 360 degrees
3sin^2 theta + 2sin theta = 0
Found 3 solutions by MathLover1, Theo, ikleyn:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
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Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
equation is 3sin^2(theta) + 2sin(theta) = 0
factor out sin(theta) to get:
sin(theta) * (3sin(theta) + 2) = 0
this is true when sin(theta) = 0 or 3sin(theta) + 2 = 0
solve for sin(theta) to get:
sin(theta) = 0 or sin(theta) = -2/3.
the reference angle (equivalent angle in the first quadrant) will be:
theta = arcsin(0) = 0 degrees.
theta = arcsin(2/3) = 41.8103149 degrees.
note that the trig function is always positive in the first quadrant.
sin(theta) is 0 when theta = 0 or 180 degrees or 360 degrees.
sin(theta) is negative in the third and fourth quadrants.
the reference angle of 41.8103149 in the third and fourth quadrant is found as follows:
third quadrant = 41.8103149 + 180 = 221.8103149 degrees.
fourth quadrant = 360 - 41.8103149 = 318.1896851 degrees.
your angles that satisfy the equation of 3sin^2(theta) + 2sin(theta) = 0 are:
0,180,360 degrees.
221.8103149, 318.1896851 degrees.
a graph of the equation is shown below:
Answer by ikleyn(52803) (Show Source): You can put this solution on YOUR website!
.
Comparing the solutions by @Theo and @MathLover1, notice that the solution by @Theo is correct.
The solution by @MathLover1 is W R O N G, showing lack of knowledge on elementary trigonometric functions.
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