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Popular >

sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi

Solution

sin^{3} (x) + 3 sin^{2} (x) + 2 sin (x) = 0, 0 \leq x < 2 π

Solution

x = 0, x = π, x = \frac{3 π}{2}

+1

Solution steps

sin^{3} (x) + 3 sin^{2} (x) + 2 sin (x) = 0, 0 \leq x < 2 π

Solve by substitution

sin (x) = 0, sin (x) = - 1, sin (x) = - 2

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

sin (x) = - 2, 0 \leq x < 2 π :

No Solution

Combine all the solutions

x = 0, x = π, x = \frac{3 π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi ?
The general solution for sin^3(x)+3sin^2(x)+2sin(x)=0,0<= x<2pi is x=0,x=pi,x=(3pi)/2