Determine solutions for the equation 5 \sin^{2} x + 2 \si

Talamancoeb

Talamancoeb

Answered question

2021-12-17

Determine solutions for the equation 5sin2x+2sinx=0 in the interval x=[2π,2π] algebraically.
Please round your irrational answers to 3 digits after the decimal point (for example x=2.378 or x=0.001). Please use appropriate notation for your final answer
please give me all solutions between [2π,2π] for this equation.

Answer & Explanation

encolatgehu

encolatgehu

Beginner2021-12-18Added 27 answers

The given trigonometric equation is 5sin2x+2sinx=0 in the interval x[2π,2π]
We have, 5sin2x+2sinx=0
sinx(5sinx+2)=0
sinx=0 or 5sinx+2=0
x=2π,π,0,2π or sinx=25
x=2π,π,0,π,2π or x=0.421,2.73,3.553,5.872
x=6.283,3.142,0,3.142,6.283 or x=0.421,2.73,3.553,5.872
x{6.283,3.142,2.73,0.421,0,3.142,3.553,5.872,6.283}
Hence all solution between [2π,2π] are x=6.283,3.142,2.730,0.421,0,3.142,3.553,5.872,6.283.
usumbiix

usumbiix

Beginner2021-12-19Added 33 answers

The solution of the equation given by;
5sin2x+2sinx=0
Is the value of the argument of sin function (x) in the question, for which the the equation is satisfied.
Now to solve the above equation, first subtract 2 sin(x) from both side of the equation as follows;
5sin2x+2sinx=0
5sin2x+2sinx2sinx=02sinx
5sin2x=2sinx (1)
sinx=0
x=sin10
=0,±π,±,2π
Now dividing both sides of the equation (1) with 5 sin(x) and solving for x as follows;
5sin2x=2sinx(1)÷(5sinx)
5sin2x5sinx=2sinx5sinx
sinx=0.4
x=sin1(0.4)
=0.412c
sin(x+2π)=sinx
x=0.412+2π
=5.871c
x=0.412c,5.871c
Therefore the solution of the equation in the interval [2π,2π] is given by;
x=0,±π,±2π,0.412,5.871 radians
Don Sumner

Don Sumner

Skilled2021-12-28Added 184 answers

5sin2x+2sinx=0
sinx[5sinx+2]=0
sinx=0
x=2π,π,0
π,2π
5sinx+2=0
sinx=25
x=sin125
x=sin125
x=2.73,0412,3.553,5.872
No of solution =9

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