What is the general solution for 4cos^2(x)+9sin(x)-9=0 ?

The general solution for 4cos^2(x)+9sin(x)-9=0 is x= pi/2+2pin

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Popular >

4cos^2(x)+9sin(x)-9=0

Solution

4 cos^{2} (x) + 9 sin (x) - 9 = 0

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) + 9 sin (x) - 9 = 0

Rewrite using trig identities

- 9 + 4 cos^{2} (x) + 9 sin (x)

Use the Pythagorean identity:

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

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

= - 9 + 4 (1 - sin^{2} (x)) + 9 sin (x)

Simplify

- 9 + 4 (1 - sin^{2} (x)) + 9 sin (x) : 9 sin (x) - 4 sin^{2} (x) - 5

- 9 + 4 (1 - sin^{2} (x)) + 9 sin (x)

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 9 + 4 - 4 sin^{2} (x) + 9 sin (x)

Add/Subtract the numbers:

- 9 + 4 = - 5

= 9 sin (x) - 4 sin^{2} (x) - 5

= 9 sin (x) - 4 sin^{2} (x) - 5

- 5 - 4 sin^{2} (x) + 9 sin (x) = 0

Solve by substitution

- 5 - 4 sin^{2} (x) + 9 sin (x) = 0

Let:

sin (x) = u

- 5 - 4 u^{2} + 9 u = 0

- 5 - 4 u^{2} + 9 u = 0 : u = 1, u = \frac{5}{4}

- 5 - 4 u^{2} + 9 u = 0

Write in the standard form

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

- 4 u^{2} + 9 u - 5 = 0

Solve with the quadratic formula

- 4 u^{2} + 9 u - 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = 9, c = - 5

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 4 ) ( - 5 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 4 ) ( - 5 )}{2 ( - 4 )}

9^{2} - 4 (- 4) (- 5) = 1

9^{2} - 4 (- 4) (- 5)

Apply rule

- (- a) = a

= 9^{2} - 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 9^{2} - 80

9^{2} = 81

= 81 - 80

Subtract the numbers:

81 - 80 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 9 \pm 1}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 9 + 1}{2 ( - 4 )}, u_{2} = \frac{- 9 - 1}{2 ( - 4 )}

u = \frac{- 9 + 1}{2 ( - 4 )} : 1

\frac{- 9 + 1}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 + 1}{- 2 \cdot 4}

Add/Subtract the numbers:

- 9 + 1 = - 8

= \frac{- 8}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{- 8}

Apply the fraction rule:

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

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 9 - 1}{2 ( - 4 )} : \frac{5}{4}

\frac{- 9 - 1}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 9 - 1}{- 2 \cdot 4}

Subtract the numbers:

- 9 - 1 = - 10

= \frac{- 10}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 10}{- 8}

Apply the fraction rule:

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

= \frac{10}{8}

Cancel the common factor:

2

= \frac{5}{4}

The solutions to the quadratic equation are:

u = 1, u = \frac{5}{4}

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = \frac{5}{4}

sin (x) = 1, sin (x) = \frac{5}{4}

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = \frac{5}{4} :

No Solution

sin (x) = \frac{5}{4}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

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tan (θ) = - \frac{12}{5}, \frac{π}{2} \leq θ \leq π

cos(x)=sin(2x+pi/2),0<= x<= 2pi

cos (x) = sin (2 x + \frac{π}{2}), 0 \leq x \leq 2 π

cos(x)=-0.397

cos (x) = - 0.397

sin(2x)cos(x)+cos(2x)sin(x)=(sqrt(3))/2

sin (2 x) cos (x) + cos (2 x) sin (x) = \frac{3}{2}

cos(x)=(5sqrt(2))/(10)

cos (x) = \frac{5 2}{10}

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+9sin(x)-9=0 ?
The general solution for 4cos^2(x)+9sin(x)-9=0 is x= pi/2+2pin