What is the general solution for 5-7sin(A)-2cos^2(A)=0 ?

The general solution for 5-7sin(A)-2cos^2(A)=0 is A= pi/6+2pin,A=(5pi)/6+2pin

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5-7sin(A)-2cos^2(A)=0

Solution

5 - 7 sin (A) - 2 cos^{2} (A) = 0

Solution

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

Degrees

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

Solution steps

5 - 7 sin (A) - 2 cos^{2} (A) = 0

Rewrite using trig identities

5 - 2 cos^{2} (A) - 7 sin (A)

Use the Pythagorean identity:

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

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

= 5 - 2 (1 - sin^{2} (A)) - 7 sin (A)

Simplify

5 - 2 (1 - sin^{2} (A)) - 7 sin (A) : 2 sin^{2} (A) - 7 sin (A) + 3

5 - 2 (1 - sin^{2} (A)) - 7 sin (A)

Expand

- 2 (1 - sin^{2} (A)) : - 2 + 2 sin^{2} (A)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (A)

= 5 - 2 + 2 sin^{2} (A) - 7 sin (A)

Subtract the numbers:

5 - 2 = 3

= 2 sin^{2} (A) - 7 sin (A) + 3

= 2 sin^{2} (A) - 7 sin (A) + 3

3 + 2 sin^{2} (A) - 7 sin (A) = 0

Solve by substitution

3 + 2 sin^{2} (A) - 7 sin (A) = 0

Let:

sin (A) = u

3 + 2 u^{2} - 7 u = 0

3 + 2 u^{2} - 7 u = 0 : u = 3, u = \frac{1}{2}

3 + 2 u^{2} - 7 u = 0

Write in the standard form

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

2 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

Subtract the numbers:

7 - 5 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 3, u = \frac{1}{2}

Substitute back

u = sin (A)

sin (A) = 3, sin (A) = \frac{1}{2}

sin (A) = 3, sin (A) = \frac{1}{2}

sin (A) = 3 :

No Solution

sin (A) = 3

- 1 \leq sin (x) \leq 1

No Solution

sin (A) = \frac{1}{2} : A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

sin (A) = \frac{1}{2}

General solutions for

sin (A) = \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}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 5-7sin(A)-2cos^2(A)=0 ?
The general solution for 5-7sin(A)-2cos^2(A)=0 is A= pi/6+2pin,A=(5pi)/6+2pin