What is the general solution for sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi ?

The general solution for sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi is x=pi+1.30899…,x=0.26179…

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sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi

Solution

2 cos (x) - 2 sin (x) = 1, 0 \leq x < 2 π

Solution

x = π + 1.30899 \dots, x = 0.26179 \dots

Degrees

x = 25 5^{\circ}, x = 1 5^{\circ}

Solution steps

2 cos (x) - 2 sin (x) = 1, 0 \leq x < 2 π

Add

2 sin (x)

to both sides

2 cos (x) = 1 + 2 sin (x)

Square both sides

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

Subtract

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

from both sides

2 cos^{2} (x) - 1 - 22 sin (x) - 2 sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 1 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 2 sin (x) 2 : - 4 sin^{2} (x) - 22 sin (x) + 1

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

= - 1 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 22 sin (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

- 1 + 2 - 2 sin^{2} (x) - 2 sin^{2} (x) - 2 sin (x) 2 : - 4 sin^{2} (x) - 22 sin (x) + 1

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

Add similar elements:

- 2 sin^{2} (x) - 2 sin^{2} (x) = - 4 sin^{2} (x)

= - 1 + 2 - 4 sin^{2} (x) - 22 sin (x)

Add/Subtract the numbers:

- 1 + 2 = 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

= - 4 sin^{2} (x) - 22 sin (x) + 1

1 - 4 sin^{2} (x) - 2 sin (x) 2 = 0

Solve by substitution

1 - 4 sin^{2} (x) - 2 sin (x) 2 = 0

Let:

sin (x) = u

1 - 4 u^{2} - 2 u 2 = 0

1 - 4 u^{2} - 2 u 2 = 0 : u = - \frac{2 + 6}{4}, u = \frac{6 - 2}{4}

1 - 4 u^{2} - 2 u 2 = 0

Write in the standard form

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

- 4 u^{2} - 22 u + 1 = 0

Solve with the quadratic formula

- 4 u^{2} - 22 u + 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 22, c = 1

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

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

(- 22)^{2} - 4 (- 4) \cdot 1 = 26

(- 22)^{2} - 4 (- 4) \cdot 1

Apply rule

- (- a) = a

= (- 22)^{2} + 4 \cdot 4 \cdot 1

(- 22)^{2} = 2^{3}

(- 22)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 22)^{2} = (22)^{2}

= (22)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (2)^{2}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 2^{2} \cdot 2

Apply exponent rule:

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

2^{2} \cdot 2 = 2^{2 + 1}

= 2^{2 + 1}

Add the numbers:

2 + 1 = 3

= 2^{3}

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{3} + 16

2^{3} = 8

= 8 + 16

Add the numbers:

8 + 16 = 24

= 24

Prime factorization of

24 : 2^{3} \cdot 3

24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

Apply exponent rule:

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

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

Apply radical rule:

= 2^{2} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 3

Refine

= 26

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

Separate the solutions

u_{1} = \frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )}

u = \frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )} : - \frac{2 + 6}{4}

\frac{- ( - 2 2 ) + 2 6}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 2 + 2 6}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

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

Apply the fraction rule:

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

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

Cancel

\frac{2 2 + 2 6}{8} : \frac{2 + 6}{4}

\frac{2 2 + 2 6}{8}

Factor out common term

2

= \frac{2 ( 2 + 6 )}{8}

Cancel the common factor:

2

= \frac{2 + 6}{4}

= - \frac{2 + 6}{4}

u = \frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )} : \frac{6 - 2}{4}

\frac{- ( - 2 2 ) - 2 6}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 2 - 2 6}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

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

Apply the fraction rule:

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

22 - 26 = - (26 - 22)

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

Factor out common term

2

= \frac{2 ( 6 - 2 )}{8}

Cancel the common factor:

2

= \frac{6 - 2}{4}

The solutions to the quadratic equation are:

u = - \frac{2 + 6}{4}, u = \frac{6 - 2}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{2 + 6}{4}, sin (x) = \frac{6 - 2}{4}

sin (x) = - \frac{2 + 6}{4}, sin (x) = \frac{6 - 2}{4}

sin (x) = - \frac{2 + 6}{4}, 0 \leq x < 2 π : x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π

sin (x) = - \frac{2 + 6}{4}, 0 \leq x < 2 π

Apply trig inverse properties

sin (x) = - \frac{2 + 6}{4}

General solutions for

sin (x) = - \frac{2 + 6}{4}

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

x = arcsin (- \frac{2 + 6}{4}) + 2 πn, x = π + arcsin (\frac{2 + 6}{4}) + 2 πn

x = arcsin (- \frac{2 + 6}{4}) + 2 πn, x = π + arcsin (\frac{2 + 6}{4}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π

sin (x) = \frac{6 - 2}{4}, 0 \leq x < 2 π : x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

sin (x) = \frac{6 - 2}{4}, 0 \leq x < 2 π

Apply trig inverse properties

sin (x) = \frac{6 - 2}{4}

General solutions for

sin (x) = \frac{6 - 2}{4}

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

x = arcsin (\frac{6 - 2}{4}) + 2 πn, x = π - arcsin (\frac{6 - 2}{4}) + 2 πn

x = arcsin (\frac{6 - 2}{4}) + 2 πn, x = π - arcsin (\frac{6 - 2}{4}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

Combine all the solutions

x = π + arcsin (\frac{2 + 6}{4}), x = - arcsin (\frac{2 + 6}{4}) + 2 π, x = arcsin (\frac{6 - 2}{4}), x = π - arcsin (\frac{6 - 2}{4})

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 cos (x) - 2 sin (x) = 1

Remove the ones that don't agree with the equation.

Check the solution

π + arcsin (\frac{2 + 6}{4}) :

True

π + arcsin (\frac{2 + 6}{4})

Plug in

n = 1

π + arcsin (\frac{2 + 6}{4})

For

2 cos (x) - 2 sin (x) = 1

plug in

x = π + arcsin (\frac{2 + 6}{4})

2 cos (π + arcsin (\frac{2 + 6}{4})) - 2 sin (π + arcsin (\frac{2 + 6}{4})) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

- arcsin (\frac{2 + 6}{4}) + 2 π :

False

- arcsin (\frac{2 + 6}{4}) + 2 π

Plug in

n = 1

- arcsin (\frac{2 + 6}{4}) + 2 π

For

2 cos (x) - 2 sin (x) = 1

plug in

x = - arcsin (\frac{2 + 6}{4}) + 2 π

2 cos (- arcsin (\frac{2 + 6}{4}) + 2 π) - 2 sin (- arcsin (\frac{2 + 6}{4}) + 2 π) = 1

Refine

1.73205 \dots = 1

\Rightarrow False

Check the solution

arcsin (\frac{6 - 2}{4}) :

True

arcsin (\frac{6 - 2}{4})

Plug in

n = 1

arcsin (\frac{6 - 2}{4})

For

2 cos (x) - 2 sin (x) = 1

plug in

x = arcsin (\frac{6 - 2}{4})

2 cos (arcsin (\frac{6 - 2}{4})) - 2 sin (arcsin (\frac{6 - 2}{4})) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

π - arcsin (\frac{6 - 2}{4}) :

False

π - arcsin (\frac{6 - 2}{4})

Plug in

n = 1

π - arcsin (\frac{6 - 2}{4})

For

2 cos (x) - 2 sin (x) = 1

plug in

x = π - arcsin (\frac{6 - 2}{4})

2 cos (π - arcsin (\frac{6 - 2}{4})) - 2 sin (π - arcsin (\frac{6 - 2}{4})) = 1

Refine

- 1.73205 \dots = 1

\Rightarrow False

x = π + arcsin (\frac{2 + 6}{4}), x = arcsin (\frac{6 - 2}{4})

Show solutions in decimal form

x = π + 1.30899 \dots, x = 0.26179 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi ?
The general solution for sqrt(2)cos(x)-sqrt(2)sin(x)=1,0<= x<2pi is x=pi+1.30899…,x=0.26179…