What is the general solution for cos(2x)=8-15sin(x) ?

The general solution for cos(2x)=8-15sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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cos(2x)=8-15sin(x)

Solution

cos (2 x) = 8 - 15 sin (x)

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

cos (2 x) = 8 - 15 sin (x)

Subtract

8 - 15 sin (x)

from both sides

cos (2 x) - 8 + 15 sin (x) = 0

Rewrite using trig identities

- 8 + cos (2 x) + 15 sin (x)

Use the Double Angle identity:

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

= - 8 + 1 - 2 sin^{2} (x) + 15 sin (x)

Simplify

= 15 sin (x) - 2 sin^{2} (x) - 7

- 7 + 15 sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

- 7 + 15 sin (x) - 2 sin^{2} (x) = 0

Let:

sin (x) = u

- 7 + 15 u - 2 u^{2} = 0

- 7 + 15 u - 2 u^{2} = 0 : u = \frac{1}{2}, u = 7

- 7 + 15 u - 2 u^{2} = 0

Write in the standard form

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

- 2 u^{2} + 15 u - 7 = 0

Solve with the quadratic formula

- 2 u^{2} + 15 u - 7 = 0

Quadratic Equation Formula:

For

a = - 2, b = 15, c = - 7

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 ( - 2 ) ( - 7 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 ( - 2 ) ( - 7 )}{2 ( - 2 )}

1 5^{2} - 4 (- 2) (- 7) = 13

1 5^{2} - 4 (- 2) (- 7)

Apply rule

- (- a) = a

= 1 5^{2} - 4 \cdot 2 \cdot 7

Multiply the numbers:

4 \cdot 2 \cdot 7 = 56

= 1 5^{2} - 56

1 5^{2} = 225

= 225 - 56

Subtract the numbers:

225 - 56 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 15 \pm 13}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 15 + 13}{2 ( - 2 )}, u_{2} = \frac{- 15 - 13}{2 ( - 2 )}

u = \frac{- 15 + 13}{2 ( - 2 )} : \frac{1}{2}

\frac{- 15 + 13}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 15 + 13}{- 2 \cdot 2}

Add/Subtract the numbers:

- 15 + 13 = - 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{- 15 - 13}{2 ( - 2 )} : 7

\frac{- 15 - 13}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 15 - 13}{- 2 \cdot 2}

Subtract the numbers:

- 15 - 13 = - 28

= \frac{- 28}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 28}{- 4}

Apply the fraction rule:

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

= \frac{28}{4}

Divide the numbers:

\frac{28}{4} = 7

= 7

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 7

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = 7

sin (x) = \frac{1}{2}, sin (x) = 7

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) = 7 :

No Solution

sin (x) = 7

- 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 cos(2x)=8-15sin(x) ?
The general solution for cos(2x)=8-15sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin