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

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

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Popular Trigonometry >

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

Solution

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

Solution

x = πn + \frac{5 π}{8}, x = πn + \frac{π}{8}

Degrees

x = 112. 5^{\circ} + 18 0^{\circ} n, x = 22. 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Factor

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

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

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

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

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

Solving each part separately

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

sin (2 x) + cos (2 x) + 2 = 0 : x = πn + \frac{5 π}{8}

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

Rewrite using trig identities

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

sin (2 x) + cos (2 x) = 2 sin (2 x + \frac{π}{4})

sin (2 x) + cos (2 x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 x) + sin (\frac{π}{4}) cos (2 x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 x + \frac{π}{4})

= 2 + 2 sin (2 x + \frac{π}{4})

2 + 2 sin (2 x + \frac{π}{4}) = 0

Move

2

to the right side

2 + 2 sin (2 x + \frac{π}{4}) = 0

Subtract

2

from both sides

2 + 2 sin (2 x + \frac{π}{4}) - 2 = 0 - 2

Simplify

2 sin (2 x + \frac{π}{4}) = - 2

2 sin (2 x + \frac{π}{4}) = - 2

Divide both sides by

2

2 sin (2 x + \frac{π}{4}) = - 2

Divide both sides by

2

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

Simplify

sin (2 x + \frac{π}{4}) = - 1

sin (2 x + \frac{π}{4}) = - 1

General solutions for

sin (2 x + \frac{π}{4}) = - 1

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}

2 x + \frac{π}{4} = \frac{3 π}{2} + 2 πn

2 x + \frac{π}{4} = \frac{3 π}{2} + 2 πn

Solve

2 x + \frac{π}{4} = \frac{3 π}{2} + 2 πn : x = πn + \frac{5 π}{8}

2 x + \frac{π}{4} = \frac{3 π}{2} + 2 πn

Move

\frac{π}{4}

to the right side

2 x + \frac{π}{4} = \frac{3 π}{2} + 2 πn

Subtract

\frac{π}{4}

from both sides

2 x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{2} + 2 πn - \frac{π}{4}

Simplify

2 x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{2} + 2 πn - \frac{π}{4}

Simplify

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

Simplify

\frac{3 π}{2} + 2 πn - \frac{π}{4} : 2 πn + \frac{5 π}{4}

\frac{3 π}{2} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{3 π}{2}

Least Common Multiplier of

4, 2 : 4

4, 2

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

4

2

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{3 π}{2} :

multiply the denominator and numerator by

2

\frac{3 π}{2} = \frac{3 π 2}{2 \cdot 2} = \frac{6 π}{4}

= - \frac{π}{4} + \frac{6 π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + 6 π}{4}

Add similar elements:

- π + 6 π = 5 π

= 2 πn + \frac{5 π}{4}

2 x = 2 πn + \frac{5 π}{4}

2 x = 2 πn + \frac{5 π}{4}

2 x = 2 πn + \frac{5 π}{4}

Divide both sides by

2

2 x = 2 πn + \frac{5 π}{4}

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{5 π}{4}}{2}

Simplify

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{5 π}{4}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{2 πn}{2} + \frac{\frac{5 π}{4}}{2} : πn + \frac{5 π}{8}

\frac{2 πn}{2} + \frac{\frac{5 π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 π}{8}

= πn + \frac{5 π}{8}

x = πn + \frac{5 π}{8}

x = πn + \frac{5 π}{8}

x = πn + \frac{5 π}{8}

x = πn + \frac{5 π}{8}

sin (2 x) + cos (2 x) - 2 = 0 : x = πn + \frac{π}{8}

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

Rewrite using trig identities

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

sin (2 x) + cos (2 x) = 2 sin (2 x + \frac{π}{4})

sin (2 x) + cos (2 x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 x) + sin (\frac{π}{4}) cos (2 x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 x + \frac{π}{4})

= - 2 + 2 sin (2 x + \frac{π}{4})

- 2 + 2 sin (2 x + \frac{π}{4}) = 0

Move

2

to the right side

- 2 + 2 sin (2 x + \frac{π}{4}) = 0

Add

2

to both sides

- 2 + 2 sin (2 x + \frac{π}{4}) + 2 = 0 + 2

Simplify

2 sin (2 x + \frac{π}{4}) = 2

2 sin (2 x + \frac{π}{4}) = 2

Divide both sides by

2

2 sin (2 x + \frac{π}{4}) = 2

Divide both sides by

2

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

Simplify

sin (2 x + \frac{π}{4}) = 1

sin (2 x + \frac{π}{4}) = 1

General solutions for

sin (2 x + \frac{π}{4}) = 1

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}

2 x + \frac{π}{4} = \frac{π}{2} + 2 πn

2 x + \frac{π}{4} = \frac{π}{2} + 2 πn

Solve

2 x + \frac{π}{4} = \frac{π}{2} + 2 πn : x = πn + \frac{π}{8}

2 x + \frac{π}{4} = \frac{π}{2} + 2 πn

Move

\frac{π}{4}

to the right side

2 x + \frac{π}{4} = \frac{π}{2} + 2 πn

Subtract

\frac{π}{4}

from both sides

2 x + \frac{π}{4} - \frac{π}{4} = \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

2 x + \frac{π}{4} - \frac{π}{4} = \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

Simplify

\frac{π}{2} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{4}

\frac{π}{2} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{2} - \frac{π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{4} - \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π}{4}

Add similar elements:

2 π - π = π

= 2 πn + \frac{π}{4}

2 x = 2 πn + \frac{π}{4}

2 x = 2 πn + \frac{π}{4}

2 x = 2 πn + \frac{π}{4}

Divide both sides by

2

2 x = 2 πn + \frac{π}{4}

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{π}{4}}{2}

Simplify

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{π}{4}}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{2 πn}{2} + \frac{\frac{π}{4}}{2} : πn + \frac{π}{8}

\frac{2 πn}{2} + \frac{\frac{π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= πn + \frac{π}{8}

x = πn + \frac{π}{8}

x = πn + \frac{π}{8}

x = πn + \frac{π}{8}

x = πn + \frac{π}{8}

Combine all the solutions

x = πn + \frac{5 π}{8}, x = πn + \frac{π}{8}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x)+cos(2x))^2=2 ?
The general solution for (sin(2x)+cos(2x))^2=2 is x=pin+(5pi)/8 ,x=pin+pi/8