What is the general solution for sin^2(a)+2cos(a)= 49/25 ?

The general solution for sin^2(a)+2cos(a)= 49/25 is a=0.64350…+2pin,a=2pi-0.64350…+2pin

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Popular >

sin^2(a)+2cos(a)= 49/25

Solution

sin^{2} (a) + 2 cos (a) = \frac{49}{25}

Solution

a = 0.64350 \dots + 2 πn, a = 2 π - 0.64350 \dots + 2 πn

Degrees

a = 36.86989 \dots^{\circ} + 36 0^{\circ} n, a = 323.13010 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (a) + 2 cos (a) = \frac{49}{25}

Subtract

\frac{49}{25}

from both sides

sin^{2} (a) + 2 cos (a) - \frac{49}{25} = 0

Simplify

sin^{2} (a) + 2 cos (a) - \frac{49}{25} : \frac{25 sin ^{2} ( a ) + 50 cos ( a ) - 49}{25}

sin^{2} (a) + 2 cos (a) - \frac{49}{25}

Convert element to fraction:

sin^{2} (a) = \frac{sin ^{2} ( a ) 25}{25}, 2 cos (a) = \frac{2 cos ( a ) 25}{25}

= \frac{sin ^{2} ( a ) \cdot 25}{25} + \frac{2 cos ( a ) \cdot 25}{25} - \frac{49}{25}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ^{2} ( a ) \cdot 25 + 2 cos ( a ) \cdot 25 - 49}{25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{25 sin ^{2} ( a ) + 50 cos ( a ) - 49}{25}

\frac{25 sin ^{2} ( a ) + 50 cos ( a ) - 49}{25} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

25 sin^{2} (a) + 50 cos (a) - 49 = 0

Rewrite using trig identities

- 49 + 25 sin^{2} (a) + 50 cos (a)

Use the Pythagorean identity:

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

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

= - 49 + 25 (1 - cos^{2} (a)) + 50 cos (a)

Simplify

- 49 + 25 (1 - cos^{2} (a)) + 50 cos (a) : 50 cos (a) - 25 cos^{2} (a) - 24

- 49 + 25 (1 - cos^{2} (a)) + 50 cos (a)

Expand

25 (1 - cos^{2} (a)) : 25 - 25 cos^{2} (a)

25 (1 - cos^{2} (a))

Apply the distributive law:

a (b - c) = ab - a c

a = 25, b = 1, c = cos^{2} (a)

= 25 \cdot 1 - 25 cos^{2} (a)

Multiply the numbers:

25 \cdot 1 = 25

= 25 - 25 cos^{2} (a)

= - 49 + 25 - 25 cos^{2} (a) + 50 cos (a)

Add/Subtract the numbers:

- 49 + 25 = - 24

= 50 cos (a) - 25 cos^{2} (a) - 24

= 50 cos (a) - 25 cos^{2} (a) - 24

- 24 - 25 cos^{2} (a) + 50 cos (a) = 0

Solve by substitution

- 24 - 25 cos^{2} (a) + 50 cos (a) = 0

Let:

cos (a) = u

- 24 - 25 u^{2} + 50 u = 0

- 24 - 25 u^{2} + 50 u = 0 : u = \frac{4}{5}, u = \frac{6}{5}

- 24 - 25 u^{2} + 50 u = 0

Write in the standard form

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

- 25 u^{2} + 50 u - 24 = 0

Solve with the quadratic formula

- 25 u^{2} + 50 u - 24 = 0

Quadratic Equation Formula:

For

a = - 25, b = 50, c = - 24

u_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 25 ) ( - 24 )}{2 ( - 25 )}

u_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 25 ) ( - 24 )}{2 ( - 25 )}

5 0^{2} - 4 (- 25) (- 24) = 10

5 0^{2} - 4 (- 25) (- 24)

Apply rule

- (- a) = a

= 5 0^{2} - 4 \cdot 25 \cdot 24

Multiply the numbers:

4 \cdot 25 \cdot 24 = 2400

= 5 0^{2} - 2400

5 0^{2} = 2500

= 2500 - 2400

Subtract the numbers:

2500 - 2400 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 50 \pm 10}{2 ( - 25 )}

Separate the solutions

u_{1} = \frac{- 50 + 10}{2 ( - 25 )}, u_{2} = \frac{- 50 - 10}{2 ( - 25 )}

u = \frac{- 50 + 10}{2 ( - 25 )} : \frac{4}{5}

\frac{- 50 + 10}{2 ( - 25 )}

Remove parentheses:

(- a) = - a

= \frac{- 50 + 10}{- 2 \cdot 25}

Add/Subtract the numbers:

- 50 + 10 = - 40

= \frac{- 40}{- 2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{- 40}{- 50}

Apply the fraction rule:

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

= \frac{40}{50}

Cancel the common factor:

10

= \frac{4}{5}

u = \frac{- 50 - 10}{2 ( - 25 )} : \frac{6}{5}

\frac{- 50 - 10}{2 ( - 25 )}

Remove parentheses:

(- a) = - a

= \frac{- 50 - 10}{- 2 \cdot 25}

Subtract the numbers:

- 50 - 10 = - 60

= \frac{- 60}{- 2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{- 60}{- 50}

Apply the fraction rule:

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

= \frac{60}{50}

Cancel the common factor:

10

= \frac{6}{5}

The solutions to the quadratic equation are:

u = \frac{4}{5}, u = \frac{6}{5}

Substitute back

u = cos (a)

cos (a) = \frac{4}{5}, cos (a) = \frac{6}{5}

cos (a) = \frac{4}{5}, cos (a) = \frac{6}{5}

cos (a) = \frac{4}{5} : a = arccos (\frac{4}{5}) + 2 πn, a = 2 π - arccos (\frac{4}{5}) + 2 πn

cos (a) = \frac{4}{5}

Apply trig inverse properties

cos (a) = \frac{4}{5}

General solutions for

cos (a) = \frac{4}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

a = arccos (\frac{4}{5}) + 2 πn, a = 2 π - arccos (\frac{4}{5}) + 2 πn

a = arccos (\frac{4}{5}) + 2 πn, a = 2 π - arccos (\frac{4}{5}) + 2 πn

cos (a) = \frac{6}{5} :

No Solution

cos (a) = \frac{6}{5}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

a = arccos (\frac{4}{5}) + 2 πn, a = 2 π - arccos (\frac{4}{5}) + 2 πn

Show solutions in decimal form

a = 0.64350 \dots + 2 πn, a = 2 π - 0.64350 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(a)+2cos(a)= 49/25 ?
The general solution for sin^2(a)+2cos(a)= 49/25 is a=0.64350…+2pin,a=2pi-0.64350…+2pin