What is the general solution for 5cos^2(y)+2sin(y)=0 ?

The general solution for 5cos^2(y)+2sin(y)=0 is y=-0.96106…+2pin,y=pi+0.96106…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5cos^2(y)+2sin(y)=0

Solution

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

Solution

y = - 0.96106 \dots + 2 πn, y = π + 0.96106 \dots + 2 πn

Degrees

y = - 55.06516 \dots^{\circ} + 36 0^{\circ} n, y = 235.06516 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

2 sin (y) + 5 cos^{2} (y)

Use the Pythagorean identity:

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

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

= 2 sin (y) + 5 (1 - sin^{2} (y))

(1 - sin^{2} (y)) \cdot 5 + 2 sin (y) = 0

Solve by substitution

(1 - sin^{2} (y)) \cdot 5 + 2 sin (y) = 0

Let:

sin (y) = u

(1 - u^{2}) \cdot 5 + 2 u = 0

(1 - u^{2}) \cdot 5 + 2 u = 0 : u = - \frac{- 1 + 26}{5}, u = \frac{1 + 26}{5}

(1 - u^{2}) \cdot 5 + 2 u = 0

Expand

(1 - u^{2}) \cdot 5 + 2 u : 5 - 5 u^{2} + 2 u

(1 - u^{2}) \cdot 5 + 2 u

= 5 (1 - u^{2}) + 2 u

Expand

5 (1 - u^{2}) : 5 - 5 u^{2}

5 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = u^{2}

= 5 \cdot 1 - 5 u^{2}

Multiply the numbers:

5 \cdot 1 = 5

= 5 - 5 u^{2}

= 5 - 5 u^{2} + 2 u

5 - 5 u^{2} + 2 u = 0

Write in the standard form

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

- 5 u^{2} + 2 u + 5 = 0

Solve with the quadratic formula

- 5 u^{2} + 2 u + 5 = 0

Quadratic Equation Formula:

For

a = - 5, b = 2, c = 5

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

2^{2} - 4 (- 5) \cdot 5 = 226

2^{2} - 4 (- 5) \cdot 5

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 2^{2} + 100

2^{2} = 4

= 4 + 100

Add the numbers:

4 + 100 = 104

= 104

Prime factorization of

104 : 2^{3} \cdot 13

104

104

divides by

2104 = 52 \cdot 2

= 2 \cdot 52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 13

= 2^{3} \cdot 13

= 2^{3} \cdot 13

Apply exponent rule:

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

= 2^{2} \cdot 2 \cdot 13

Apply radical rule:

= 2^{2} 2 \cdot 13

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 13

Refine

= 226

u_{1, 2} = \frac{- 2 \pm 2 26}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- 2 + 2 26}{2 ( - 5 )}, u_{2} = \frac{- 2 - 2 26}{2 ( - 5 )}

u = \frac{- 2 + 2 26}{2 ( - 5 )} : - \frac{- 1 + 26}{5}

\frac{- 2 + 2 26}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 26}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 2 + 2 26}{- 10}

Apply the fraction rule:

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

= - \frac{- 2 + 2 26}{10}

Cancel

\frac{- 2 + 2 26}{10} : \frac{26 - 1}{5}

\frac{- 2 + 2 26}{10}

Factor

- 2 + 226 : 2 (- 1 + 26)

- 2 + 226

Rewrite as

= - 2 \cdot 1 + 226

Factor out common term

2

= 2 (- 1 + 26)

= \frac{2 ( - 1 + 26 )}{10}

Cancel the common factor:

2

= \frac{- 1 + 26}{5}

= - \frac{26 - 1}{5}

= - \frac{- 1 + 26}{5}

u = \frac{- 2 - 2 26}{2 ( - 5 )} : \frac{1 + 26}{5}

\frac{- 2 - 2 26}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 26}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 2 - 2 26}{- 10}

Apply the fraction rule:

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

- 2 - 226 = - (2 + 226)

= \frac{2 + 2 26}{10}

Factor

2 + 226 : 2 (1 + 26)

2 + 226

Rewrite as

= 2 \cdot 1 + 226

Factor out common term

2

= 2 (1 + 26)

= \frac{2 ( 1 + 26 )}{10}

Cancel the common factor:

2

= \frac{1 + 26}{5}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 26}{5}, u = \frac{1 + 26}{5}

Substitute back

u = sin (y)

sin (y) = - \frac{- 1 + 26}{5}, sin (y) = \frac{1 + 26}{5}

sin (y) = - \frac{- 1 + 26}{5}, sin (y) = \frac{1 + 26}{5}

sin (y) = - \frac{- 1 + 26}{5} : y = arcsin (- \frac{- 1 + 26}{5}) + 2 πn, y = π + arcsin (\frac{- 1 + 26}{5}) + 2 πn

sin (y) = - \frac{- 1 + 26}{5}

Apply trig inverse properties

sin (y) = - \frac{- 1 + 26}{5}

General solutions for

sin (y) = - \frac{- 1 + 26}{5}

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

y = arcsin (- \frac{- 1 + 26}{5}) + 2 πn, y = π + arcsin (\frac{- 1 + 26}{5}) + 2 πn

y = arcsin (- \frac{- 1 + 26}{5}) + 2 πn, y = π + arcsin (\frac{- 1 + 26}{5}) + 2 πn

sin (y) = \frac{1 + 26}{5} :

No Solution

sin (y) = \frac{1 + 26}{5}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

y = arcsin (- \frac{- 1 + 26}{5}) + 2 πn, y = π + arcsin (\frac{- 1 + 26}{5}) + 2 πn

Show solutions in decimal form

y = - 0.96106 \dots + 2 πn, y = π + 0.96106 \dots + 2 πn

Popular Examples

tan(θ)=(sqrt(2))/2

tan (θ) = \frac{2}{2}

tan^2(θ)-16=0

tan^{2} (θ) - 16 = 0

(0.5)(9.8)=(0.45)(9.8)sin(θ)

(0.5) (9.8) = (0.45) (9.8) sin (θ)

tan(θ)=-1.732

tan (θ) = - 1.732

sin(θ/2)+cos(θ)=0

sin (\frac{θ}{2}) + cos (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(y)+2sin(y)=0 ?
The general solution for 5cos^2(y)+2sin(y)=0 is y=-0.96106…+2pin,y=pi+0.96106…+2pin