What is the general solution for 2sin(x)=2+cos(2x) ?

The general solution for 2sin(x)=2+cos(2x) is x=0.96645…+2pin,x=pi-0.96645…+2pin

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2sin(x)=2+cos(2x)

Solution

2 sin (x) = 2 + cos (2 x)

Solution

x = 0.96645 \dots + 2 πn, x = π - 0.96645 \dots + 2 πn

Degrees

x = 55.37370 \dots^{\circ} + 36 0^{\circ} n, x = 124.62629 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) = 2 + cos (2 x)

Subtract

2 + cos (2 x)

from both sides

2 sin (x) - 2 - cos (2 x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Subtract the numbers:

- 2 - 1 = - 3

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 \cdot 2 (- 3) = 27

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 2^{2} + 24

2^{2} = 4

= 4 + 24

Add the numbers:

4 + 24 = 28

= 28

Prime factorization of

28 : 2^{2} \cdot 7

28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

Apply radical rule:

= 7 2^{2}

Apply radical rule:

2^{2} = 2

= 27

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

Separate the solutions

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

u = \frac{- 2 + 2 7}{2 \cdot 2} : \frac{- 1 + 7}{2}

\frac{- 2 + 2 7}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 7}{4}

Factor

- 2 + 27 : 2 (- 1 + 7)

- 2 + 27

Rewrite as

= - 2 \cdot 1 + 27

Factor out common term

2

= 2 (- 1 + 7)

= \frac{2 ( - 1 + 7 )}{4}

Cancel the common factor:

2

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

u = \frac{- 2 - 2 7}{2 \cdot 2} : - \frac{1 + 7}{2}

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 7}{4}

Factor

- 2 - 27 : - 2 (1 + 7)

- 2 - 27

Rewrite as

= - 2 \cdot 1 - 27

Factor out common term

2

= - 2 (1 + 7)

= - \frac{2 ( 1 + 7 )}{4}

Cancel the common factor:

2

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

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{- 1 + 7}{2} : x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

sin (x) = - \frac{1 + 7}{2} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

Show solutions in decimal form

x = 0.96645 \dots + 2 πn, x = π - 0.96645 \dots + 2 πn

Popular Examples

sin(x)=(-4/3 sqrt(3))/(8/3 sqrt(3))

sin (x) = \frac{- \frac{4}{3} 3}{\frac{8}{3} 3}

sec(a)= 5/4

sec (a) = \frac{5}{4}

cot^4(x)-9=0

cot^{4} (x) - 9 = 0

0.5*sin(x)=1*(645*10^{-9})

0.5 \cdot sin (x) = 1 \cdot (645 \cdot 1 0^{- 9})

sin(2x)=2cos(2x)

sin (2 x) = 2 cos (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)=2+cos(2x) ?
The general solution for 2sin(x)=2+cos(2x) is x=0.96645…+2pin,x=pi-0.96645…+2pin