What is the general solution for 2cos^2(x)+5sin(x)-4=0 ?

The general solution for 2cos^2(x)+5sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2cos^2(x)+5sin(x)-4=0

Solution

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

Solution

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

Degrees

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

Solution steps

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

Rewrite using trig identities

- 4 + 2 cos^{2} (x) + 5 sin (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Add/Subtract the numbers:

- 4 + 2 = - 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

5^{2} - 4 (- 2) (- 2) = 3

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 5 + 3}{2 ( - 2 )}, u_{2} = \frac{- 5 - 3}{2 ( - 2 )}

u = \frac{- 5 + 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- 5 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 5 + 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 5 - 3}{2 ( - 2 )} : 2

\frac{- 5 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 5 - 3 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{- 4}

Apply the fraction rule:

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

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+5sin(x)-4=0 ?
The general solution for 2cos^2(x)+5sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin