What is the general solution for solvefor x,sin(x)=a-2*q*cos(2x) ?

The general solution for solvefor x,sin(x)=a-2*q*cos(2x) is

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solvefor x,sin(x)=a-2qcos(2x)

Solution

solve for

x, sin (x) = a - 2 \cdot q \cdot cos (2 x)

Solution

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

Solution steps

sin (x) = a - 2 q cos (2 x)

Subtract

a - 2 q cos (2 x)

from both sides

sin (x) - a + 2 q cos (2 x) = 0

Rewrite using trig identities

sin (x) - a + 2 cos (2 x) q

Use the Double Angle identity:

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

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

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

Solve by substitution

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

Let:

sin (x) = u

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0 : u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

Expand

u - a + (1 - 2 u^{2}) \cdot 2 q : u - a + 2 q - 4 q u^{2}

u - a + (1 - 2 u^{2}) \cdot 2 q

= u - a + 2 q (1 - 2 u^{2})

Expand

2 q (1 - 2 u^{2}) : 2 q - 4 q u^{2}

2 q (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2 q, b = 1, c = 2 u^{2}

= 2 q \cdot 1 - 2 q \cdot 2 u^{2}

= 2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

Simplify

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2} : 2 q - 4 q u^{2}

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 q - 2 \cdot 2 q u^{2}

Multiply the numbers:

2 \cdot 2 = 4

= 2 q - 4 q u^{2}

= 2 q - 4 q u^{2}

= u - a + 2 q - 4 q u^{2}

u - a + 2 q - 4 q u^{2} = 0

Write in the standard form

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

- 4 q u^{2} + u - a + 2 q = 0

Solve with the quadratic formula

- 4 q u^{2} + u - a + 2 q = 0

Quadratic Equation Formula:

For

a = - 4 q, b = 1, c = - a + 2 q

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

Simplify

1^{2} - 4 (- 4 q) (- a + 2 q) : 1 + 16 q (- a + 2 q)

1^{2} - 4 (- 4 q) (- a + 2 q)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4 q) (2 q - a)

Apply rule

- (- a) = a

= 1 + 4 \cdot 4 q (- a + 2 q)

Multiply the numbers:

4 \cdot 4 = 16

= 1 + 16 q (2 q - a)

u_{1, 2} = \frac{- 1 \pm 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}; q \neq = 0

Separate the solutions

u_{1} = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}, u_{2} = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

u = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 1 + 16 q ( 2 q - a ) + 1}{- 8 q}

Apply the fraction rule:

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

= - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

u = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 1 - 16 q ( 2 q - a ) + 1}{- 8 q}

Apply the fraction rule:

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

= - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

General solutions for

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

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

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

General solutions for

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

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

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,sin(x)=a-2*q*cos(2x) ?
The general solution for solvefor x,sin(x)=a-2*q*cos(2x) is