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

The general solution for 3+sin(y)=3cos^2(y) is y=2pin,y=pi+2pin,y=-0.33983…+2pin,y=pi+0.33983…+2pin

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3+sin(y)=3cos^2(y)

Solution

3 + sin (y) = 3 cos^{2} (y)

Solution

y = 2 πn, y = π + 2 πn, y = - 0.33983 \dots + 2 πn, y = π + 0.33983 \dots + 2 πn

Degrees

y = 0^{\circ} + 36 0^{\circ} n, y = 18 0^{\circ} + 36 0^{\circ} n, y = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, y = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 + sin (y) = 3 cos^{2} (y)

Subtract

3 cos^{2} (y)

from both sides

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

Rewrite using trig identities

3 + sin (y) - 3 cos^{2} (y)

Use the Pythagorean identity:

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

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

= 3 + sin (y) - 3 (1 - sin^{2} (y))

Simplify

3 + sin (y) - 3 (1 - sin^{2} (y)) : 3 sin^{2} (y) + sin (y)

3 + sin (y) - 3 (1 - sin^{2} (y))

Expand

- 3 (1 - sin^{2} (y)) : - 3 + 3 sin^{2} (y)

- 3 (1 - sin^{2} (y))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (y)

= - 3 \cdot 1 - (- 3) sin^{2} (y)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (y)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (y)

= 3 + sin (y) - 3 + 3 sin^{2} (y)

Simplify

3 + sin (y) - 3 + 3 sin^{2} (y) : 3 sin^{2} (y) + sin (y)

3 + sin (y) - 3 + 3 sin^{2} (y)

Group like terms

= sin (y) + 3 sin^{2} (y) + 3 - 3

3 - 3 = 0

= 3 sin^{2} (y) + sin (y)

= 3 sin^{2} (y) + sin (y)

= 3 sin^{2} (y) + sin (y)

sin (y) + 3 sin^{2} (y) = 0

Solve by substitution

sin (y) + 3 sin^{2} (y) = 0

Let:

sin (y) = u

u + 3 u^{2} = 0

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

u + 3 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

3 u^{2} + u = 0

Solve with the quadratic formula

3 u^{2} + u = 0

Quadratic Equation Formula:

For

a = 3, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

1^{2} - 4 \cdot 3 \cdot 0 = 1

1^{2} - 4 \cdot 3 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 \cdot 3}, u_{2} = \frac{- 1 - 1}{2 \cdot 3}

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

\frac{- 1 + 1}{2 \cdot 3}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 1 - 1}{2 \cdot 3} : - \frac{1}{3}

\frac{- 1 - 1}{2 \cdot 3}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

The solutions to the quadratic equation are:

u = 0, u = - \frac{1}{3}

Substitute back

u = sin (y)

sin (y) = 0, sin (y) = - \frac{1}{3}

sin (y) = 0, sin (y) = - \frac{1}{3}

sin (y) = 0 : y = 2 πn, y = π + 2 πn

sin (y) = 0

General solutions for

sin (y) = 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}

y = 0 + 2 πn, y = π + 2 πn

y = 0 + 2 πn, y = π + 2 πn

Solve

y = 0 + 2 πn : y = 2 πn

y = 0 + 2 πn

0 + 2 πn = 2 πn

y = 2 πn

y = 2 πn, y = π + 2 πn

sin (y) = - \frac{1}{3} : y = arcsin (- \frac{1}{3}) + 2 πn, y = π + arcsin (\frac{1}{3}) + 2 πn

sin (y) = - \frac{1}{3}

Apply trig inverse properties

sin (y) = - \frac{1}{3}

General solutions for

sin (y) = - \frac{1}{3}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

y = arcsin (- \frac{1}{3}) + 2 πn, y = π + arcsin (\frac{1}{3}) + 2 πn

y = arcsin (- \frac{1}{3}) + 2 πn, y = π + arcsin (\frac{1}{3}) + 2 πn

Combine all the solutions

y = 2 πn, y = π + 2 πn, y = arcsin (- \frac{1}{3}) + 2 πn, y = π + arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

y = 2 πn, y = π + 2 πn, y = - 0.33983 \dots + 2 πn, y = π + 0.33983 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3+sin(y)=3cos^2(y) ?
The general solution for 3+sin(y)=3cos^2(y) is y=2pin,y=pi+2pin,y=-0.33983…+2pin,y=pi+0.33983…+2pin