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

The general solution for 5cos^2(a)-2sin(a)-2=0 is a=(3pi)/2+2pin,a=0.64350…+2pin,a=pi-0.64350…+2pin

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Popular >

5cos^2(a)-2sin(a)-2=0

Solution

5 cos^{2} (a) - 2 sin (a) - 2 = 0

Solution

a = \frac{3 π}{2} + 2 πn, a = 0.64350 \dots + 2 πn, a = π - 0.64350 \dots + 2 πn

Degrees

a = 27 0^{\circ} + 36 0^{\circ} n, a = 36.86989 \dots^{\circ} + 36 0^{\circ} n, a = 143.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos^{2} (a) - 2 sin (a) - 2 = 0

Rewrite using trig identities

- 2 - 2 sin (a) + 5 cos^{2} (a)

Use the Pythagorean identity:

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

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

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

Simplify

- 2 - 2 sin (a) + 5 (1 - sin^{2} (a)) : - 5 sin^{2} (a) - 2 sin (a) + 3

- 2 - 2 sin (a) + 5 (1 - sin^{2} (a))

Expand

5 (1 - sin^{2} (a)) : 5 - 5 sin^{2} (a)

5 (1 - sin^{2} (a))

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = sin^{2} (a)

= 5 \cdot 1 - 5 sin^{2} (a)

Multiply the numbers:

5 \cdot 1 = 5

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

= - 2 - 2 sin (a) + 5 - 5 sin^{2} (a)

Simplify

- 2 - 2 sin (a) + 5 - 5 sin^{2} (a) : - 5 sin^{2} (a) - 2 sin (a) + 3

- 2 - 2 sin (a) + 5 - 5 sin^{2} (a)

Group like terms

= - 2 sin (a) - 5 sin^{2} (a) - 2 + 5

Add/Subtract the numbers:

- 2 + 5 = 3

= - 5 sin^{2} (a) - 2 sin (a) + 3

= - 5 sin^{2} (a) - 2 sin (a) + 3

= - 5 sin^{2} (a) - 2 sin (a) + 3

3 - 2 sin (a) - 5 sin^{2} (a) = 0

Solve by substitution

3 - 2 sin (a) - 5 sin^{2} (a) = 0

Let:

sin (a) = u

3 - 2 u - 5 u^{2} = 0

3 - 2 u - 5 u^{2} = 0 : u = - 1, u = \frac{3}{5}

3 - 2 u - 5 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 5, b = - 2, c = 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

(- 2)^{2} - 4 (- 5) \cdot 3 = 8

(- 2)^{2} - 4 (- 5) \cdot 3

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 5 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 2^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 8}{2 ( - 5 )}, u_{2} = \frac{- ( - 2 ) - 8}{2 ( - 5 )}

u = \frac{- ( - 2 ) + 8}{2 ( - 5 )} : - 1

\frac{- ( - 2 ) + 8}{2 ( - 5 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 + 8}{- 2 \cdot 5}

Add the numbers:

2 + 8 = 10

= \frac{10}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{- 10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 2 ) - 8}{2 ( - 5 )} : \frac{3}{5}

\frac{- ( - 2 ) - 8}{2 ( - 5 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{2 - 8}{- 2 \cdot 5}

Subtract the numbers:

2 - 8 = - 6

= \frac{- 6}{- 2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6}{- 10}

Apply the fraction rule:

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

= \frac{6}{10}

Cancel the common factor:

2

= \frac{3}{5}

The solutions to the quadratic equation are:

u = - 1, u = \frac{3}{5}

Substitute back

u = sin (a)

sin (a) = - 1, sin (a) = \frac{3}{5}

sin (a) = - 1, sin (a) = \frac{3}{5}

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

sin (a) = - 1

General solutions for

sin (a) = - 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}

a = \frac{3 π}{2} + 2 πn

a = \frac{3 π}{2} + 2 πn

sin (a) = \frac{3}{5} : a = arcsin (\frac{3}{5}) + 2 πn, a = π - arcsin (\frac{3}{5}) + 2 πn

sin (a) = \frac{3}{5}

Apply trig inverse properties

sin (a) = \frac{3}{5}

General solutions for

sin (a) = \frac{3}{5}

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

a = arcsin (\frac{3}{5}) + 2 πn, a = π - arcsin (\frac{3}{5}) + 2 πn

a = arcsin (\frac{3}{5}) + 2 πn, a = π - arcsin (\frac{3}{5}) + 2 πn

Combine all the solutions

a = \frac{3 π}{2} + 2 πn, a = arcsin (\frac{3}{5}) + 2 πn, a = π - arcsin (\frac{3}{5}) + 2 πn

Show solutions in decimal form

a = \frac{3 π}{2} + 2 πn, a = 0.64350 \dots + 2 πn, a = π - 0.64350 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(a)-2sin(a)-2=0 ?
The general solution for 5cos^2(a)-2sin(a)-2=0 is a=(3pi)/2+2pin,a=0.64350…+2pin,a=pi-0.64350…+2pin