What is the general solution for -3cos^2(u)sin(u)-2sin(2u)=0 ?

The general solution for -3cos^2(u)sin(u)-2sin(2u)=0 is u=2pin,u=pi+2pin,u= pi/2+2pin,u=(3pi)/2+2pin

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Popular >

-3cos^2(u)sin(u)-2sin(2u)=0

Solution

- 3 cos^{2} (u) sin (u) - 2 sin (2 u) = 0

Solution

u = 2 πn, u = π + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

Degrees

u = 0^{\circ} + 36 0^{\circ} n, u = 18 0^{\circ} + 36 0^{\circ} n, u = 9 0^{\circ} + 36 0^{\circ} n, u = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos^{2} (u) sin (u) - 2 sin (2 u) = 0

Rewrite using trig identities

- 2 sin (2 u) - 3 cos^{2} (u) sin (u)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 2 \cdot 2 sin (u) cos (u) - 3 cos^{2} (u) sin (u)

Simplify

= - 4 sin (u) cos (u) - 3 cos^{2} (u) sin (u)

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u) = 0

Factor

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u) : - sin (u) cos (u) (3 cos (u) + 4)

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin (u) cos^{2} (u) = sin (u) cos (u) cos (u)

= - 3 sin (u) cos (u) cos (u) - 4 sin (u) cos (u)

Factor out common term

- sin (u) cos (u)

= - sin (u) cos (u) (3 cos (u) + 4)

- sin (u) cos (u) (3 cos (u) + 4) = 0

Solving each part separately

sin (u) = 0 or cos (u) = 0 or 3 cos (u) + 4 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

General solutions for

sin (u) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

General solutions for

cos (u) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

3 cos (u) + 4 = 0 :

No Solution

3 cos (u) + 4 = 0

Move

4

to the right side

3 cos (u) + 4 = 0

Subtract

4

from both sides

3 cos (u) + 4 - 4 = 0 - 4

Simplify

3 cos (u) = - 4

3 cos (u) = - 4

Divide both sides by

3

3 cos (u) = - 4

Divide both sides by

3

\frac{3 cos ( u )}{3} = \frac{- 4}{3}

Simplify

cos (u) = - \frac{4}{3}

cos (u) = - \frac{4}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -3cos^2(u)sin(u)-2sin(2u)=0 ?
The general solution for -3cos^2(u)sin(u)-2sin(2u)=0 is u=2pin,u=pi+2pin,u= pi/2+2pin,u=(3pi)/2+2pin