What is the general solution for 2sin^3(x)+cos^2(x)=1 ?

The general solution for 2sin^3(x)+cos^2(x)=1 is x=2pin,x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^3(x)+cos^2(x)=1

Solution

2 sin^{3} (x) + cos^{2} (x) = 1

Solution

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{3} (x) + cos^{2} (x) = 1

Subtract

1

from both sides

2 sin^{3} (x) + cos^{2} (x) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (x) + 2 sin^{3} (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= 2 sin^{3} (x) - sin^{2} (x)

- sin^{2} (x) + 2 sin^{3} (x) = 0

Solve by substitution

- sin^{2} (x) + 2 sin^{3} (x) = 0

Let:

sin (x) = u

- u^{2} + 2 u^{3} = 0

- u^{2} + 2 u^{3} = 0 : u = 0, u = \frac{1}{2}

- u^{2} + 2 u^{3} = 0

Factor

- u^{2} + 2 u^{3} : u^{2} (2 u - 1)

- u^{2} + 2 u^{3}

Apply exponent rule:

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

u^{3} = u u^{2}

= 2 u u^{2} - u^{2}

Factor out common term

u^{2}

= u^{2} (2 u - 1)

u^{2} (2 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 2 u - 1 = 0

Solve

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

The solutions are

u = 0, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0, sin (x) = \frac{1}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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tan (x) = \frac{59}{56}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^3(x)+cos^2(x)=1 ?
The general solution for 2sin^3(x)+cos^2(x)=1 is x=2pin,x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin