What is the general solution for (3cos(2x)+7sin(x)-5)/(9cos^2(x)-5)=0 ?

The general solution for (3cos(2x)+7sin(x)-5)/(9cos^2(x)-5)=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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(3cos(2x)+7sin(x)-5)/(9cos^2(x)-5)=0

Solution

\frac{3 cos ( 2 x ) + 7 sin ( x ) - 5}{9 cos ^{2} ( x ) - 5} = 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

\frac{3 cos ( 2 x ) + 7 sin ( x ) - 5}{9 cos ^{2} ( x ) - 5} = 0

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

3 cos (2 x) + 7 sin (x) - 5 = 0

Rewrite using trig identities

- 5 + 3 cos (2 x) + 7 sin (x)

Use the Double Angle identity:

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

= - 5 + 3 (1 - 2 sin^{2} (x)) + 7 sin (x)

Simplify

- 5 + 3 (1 - 2 sin^{2} (x)) + 7 sin (x) : 7 sin (x) - 6 sin^{2} (x) - 2

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

Expand

3 (1 - 2 sin^{2} (x)) : 3 - 6 sin^{2} (x)

3 (1 - 2 sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 sin^{2} (x)

= 3 \cdot 1 - 3 \cdot 2 sin^{2} (x)

Simplify

3 \cdot 1 - 3 \cdot 2 sin^{2} (x) : 3 - 6 sin^{2} (x)

3 \cdot 1 - 3 \cdot 2 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 \cdot 2 sin^{2} (x)

Multiply the numbers:

3 \cdot 2 = 6

= 3 - 6 sin^{2} (x)

= 3 - 6 sin^{2} (x)

= - 5 + 3 - 6 sin^{2} (x) + 7 sin (x)

Add/Subtract the numbers:

- 5 + 3 = - 2

= 7 sin (x) - 6 sin^{2} (x) - 2

= 7 sin (x) - 6 sin^{2} (x) - 2

- 2 - 6 sin^{2} (x) + 7 sin (x) = 0

Solve by substitution

- 2 - 6 sin^{2} (x) + 7 sin (x) = 0

Let:

sin (x) = u

- 2 - 6 u^{2} + 7 u = 0

- 2 - 6 u^{2} + 7 u = 0 : u = \frac{1}{2}, u = \frac{2}{3}

- 2 - 6 u^{2} + 7 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 6, b = 7, c = - 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 6 ) ( - 2 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 6 ) ( - 2 )}{2 ( - 6 )}

7^{2} - 4 (- 6) (- 2) = 1

7^{2} - 4 (- 6) (- 2)

Apply rule

- (- a) = a

= 7^{2} - 4 \cdot 6 \cdot 2

Multiply the numbers:

4 \cdot 6 \cdot 2 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 7 \pm 1}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- 7 + 1}{2 ( - 6 )}, u_{2} = \frac{- 7 - 1}{2 ( - 6 )}

u = \frac{- 7 + 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 7 + 1}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 + 1}{- 2 \cdot 6}

Add/Subtract the numbers:

- 7 + 1 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

Apply the fraction rule:

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

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

u = \frac{- 7 - 1}{2 ( - 6 )} : \frac{2}{3}

\frac{- 7 - 1}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 - 1}{- 2 \cdot 6}

Subtract the numbers:

- 7 - 1 = - 8

= \frac{- 8}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

Apply the fraction rule:

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

= \frac{8}{12}

Cancel the common factor:

4

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (\frac{2}{3}) + 2 πn, x = π - arcsin (\frac{2}{3}) + 2 πn

Since the equation is undefined for:

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (3cos(2x)+7sin(x)-5)/(9cos^2(x)-5)=0 ?
The general solution for (3cos(2x)+7sin(x)-5)/(9cos^2(x)-5)=0 is x= pi/6+2pin,x=(5pi)/6+2pin