SOLUTION: Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate solutions to the nearest ten-thousandth.
4 sin^3 x = sin x
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Question 1178160: Solve the equation for solutions in the interval 0 ≤ x < 2𝜋. Round approximate solutions to the nearest ten-thousandth.
4 sin^3 x = sin x
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
this is 4sin^3x-sin x=0
sin x(4 sin^2x-1)=0=sin x(2 sin x+1)(2 sin x-1)
x=0 degrees
x=30 and 150 degrees
x=210 and 330 degrees; graph in radians
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
or
since given the interval, go with positive solution sine is positive in quadrants I and II only
solutions in radians:
->rounded to the nearest ten-thousandth
solutions in degrees:
°
°
°
°
°
°
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