What is the general solution for (2-2sin(x))/a = 2/(sin(x)) ?

The general solution for (2-2sin(x))/a = 2/(sin(x)) is

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(2-2sin(x))/a = 2/(sin(x))

Solution

\frac{2 - 2 sin ( x )}{a} = \frac{2}{sin ( x )}

Solution

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

Solution steps

\frac{2 - 2 sin ( x )}{a} = \frac{2}{sin ( x )}

Solve by substitution

\frac{2 - 2 sin ( x )}{a} = \frac{2}{sin ( x )}

Let:

sin (x) = u

\frac{2 - 2 u}{a} = \frac{2}{u}

\frac{2 - 2 u}{a} = \frac{2}{u} : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

\frac{2 - 2 u}{a} = \frac{2}{u}

Cross multiply

\frac{2 - 2 u}{a} = \frac{2}{u}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

(2 - 2 u) u = a \cdot 2

(2 - 2 u) u = a \cdot 2

Solve

(2 - 2 u) u = a \cdot 2 : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

(2 - 2 u) u = a \cdot 2

Expand

(2 - 2 u) u : 2 u - 2 u^{2}

(2 - 2 u) u

= u (2 - 2 u)

Apply the distributive law:

a (b - c) = ab - a c

a = u, b = 2, c = 2 u

= u \cdot 2 - u \cdot 2 u

= 2 u - 2 uu

2 uu = 2 u^{2}

2 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

= 2 u - 2 u^{2}

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

Move

a 2

to the left side

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

Subtract

a 2

from both sides

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

Simplify

2 u - 2 u^{2} - a \cdot 2 = 0

2 u - 2 u^{2} - a \cdot 2 = 0

Write in the standard form

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

- 2 u^{2} + 2 u - a \cdot 2 = 0

Solve with the quadratic formula

- 2 u^{2} + 2 u - a \cdot 2 = 0

Quadratic Equation Formula:

For

a = - 2, b = 2, c = - a 2

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

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

Simplify

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

2^{2} - 4 (- 2) (- a \cdot 2)

Apply rule

- (- a) = a

= 2^{2} - 4 \cdot 2 a \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 2^{2} - 16 a

Factor

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

2^{2} - 16 a

Rewrite as

= 4 \cdot 1 - 4 \cdot 4 a

Factor out common term

4

= 4 (1 - 4 a)

= 4 (1 - 4 a)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 - 4 a + 1

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 - 4 a + 1

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 1 - 4 a}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 - 4 a + 1}{- 4}

Apply the fraction rule:

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

= - \frac{- 2 + 2 1 - 4 a}{4}

Cancel

\frac{- 2 + 2 1 - 4 a}{4} : \frac{- 4 a + 1 - 1}{2}

\frac{- 2 + 2 1 - 4 a}{4}

Factor

- 2 + 2 1 - 4 a : 2 (- 1 + 1 - 4 a)

- 2 + 2 1 - 4 a

Rewrite as

= - 2 \cdot 1 + 2 1 - 4 a

Factor out common term

2

= 2 (- 1 + 1 - 4 a)

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

Cancel the common factor:

2

= \frac{- 1 + - 4 a + 1}{2}

= - \frac{- 4 a + 1 - 1}{2}

= - \frac{- 1 + - 4 a + 1}{2}

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

\frac{- 2 - 2 1 - 4 a}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 1 - 4 a}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 - 4 a + 1}{- 4}

Apply the fraction rule:

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

= - \frac{- 2 - 2 1 - 4 a}{4}

Cancel

\frac{- 2 - 2 1 - 4 a}{4} : - \frac{- 4 a + 1 + 1}{2}

\frac{- 2 - 2 1 - 4 a}{4}

Factor

- 2 - 2 1 - 4 a : - 2 (1 + 1 - 4 a)

- 2 - 2 1 - 4 a

Rewrite as

= - 2 \cdot 1 - 2 1 - 4 a

Factor out common term

2

= - 2 (1 + 1 - 4 a)

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

Cancel the common factor:

2

= - \frac{- 4 a + 1 + 1}{2}

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

Apply rule

- (- a) = a

= \frac{- 4 a + 1 + 1}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

sin (x) = - \frac{- 1 + - 4 a + 1}{2} : x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

General solutions for

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

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

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

sin (x) = \frac{- 4 a + 1 + 1}{2} : x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

sin (x) = \frac{- 4 a + 1 + 1}{2}

Apply trig inverse properties

sin (x) = \frac{- 4 a + 1 + 1}{2}

General solutions for

sin (x) = \frac{- 4 a + 1 + 1}{2}

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

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (2-2sin(x))/a = 2/(sin(x)) ?
The general solution for (2-2sin(x))/a = 2/(sin(x)) is