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

The general solution for sqrt(2)sin(x)-sqrt(2)cos(x)=2 is x=(3pi)/4+2pin

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Popular Trigonometry >

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

Solution

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

Solution

x = \frac{3 π}{4} + 2 πn

Degrees

x = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

2 cos (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

2 sin^{2} (x) - 4 - 42 cos (x) - 2 cos^{2} (x) = 0

Rewrite using trig identities

- 4 - 2 cos^{2} (x) + 2 sin^{2} (x) - 4 cos (x) 2

Use the Pythagorean identity:

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

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

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

Simplify

- 4 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 4 cos (x) 2 : - 4 cos^{2} (x) - 42 cos (x) - 2

- 4 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 4 cos (x) 2

= - 4 - 2 cos^{2} (x) + 2 (1 - cos^{2} (x)) - 42 cos (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

= - 4 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 4 cos (x) 2

Simplify

- 4 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 4 cos (x) 2 : - 4 cos^{2} (x) - 42 cos (x) - 2

- 4 - 2 cos^{2} (x) + 2 - 2 cos^{2} (x) - 4 cos (x) 2

Group like terms

= - 2 cos^{2} (x) - 2 cos^{2} (x) - 42 cos (x) - 4 + 2

Add similar elements:

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

= - 4 cos^{2} (x) - 42 cos (x) - 4 + 2

Add/Subtract the numbers:

- 4 + 2 = - 2

= - 4 cos^{2} (x) - 42 cos (x) - 2

= - 4 cos^{2} (x) - 42 cos (x) - 2

= - 4 cos^{2} (x) - 42 cos (x) - 2

- 2 - 4 cos^{2} (x) - 4 cos (x) 2 = 0

Solve by substitution

- 2 - 4 cos^{2} (x) - 4 cos (x) 2 = 0

Let:

cos (x) = u

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

- 2 - 4 u^{2} - 4 u 2 = 0 : u = - \frac{2}{2}

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

Write in the standard form

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

- 4 u^{2} - 42 u - 2 = 0

Solve with the quadratic formula

- 4 u^{2} - 42 u - 2 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 42, c = - 2

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

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

(- 42)^{2} - 4 (- 4) (- 2) = 0

(- 42)^{2} - 4 (- 4) (- 2)

Apply rule

- (- a) = a

= (- 42)^{2} - 4 \cdot 4 \cdot 2

(- 42)^{2} = 4^{2} \cdot 2

(- 42)^{2}

Apply exponent rule:

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

n

is even

(- 42)^{2} = (42)^{2}

= (42)^{2}

Apply exponent rule:

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

= 4^{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

= 4^{2} \cdot 2

4 \cdot 4 \cdot 2 = 32

4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 32

= 4^{2} \cdot 2 - 32

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

Multiply the numbers:

16 \cdot 2 = 32

= 32

= 32 - 32

Subtract the numbers:

32 - 32 = 0

= 0

u_{1, 2} = \frac{- ( - 4 2 ) \pm 0}{2 ( - 4 )}

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

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 2}{- 8}

Apply the fraction rule:

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

= - \frac{4 2}{8}

Cancel the common factor:

4

= - \frac{2}{2}

u = - \frac{2}{2}

The solution to the quadratic equation is:

u = - \frac{2}{2}

Substitute back

u = cos (x)

cos (x) = - \frac{2}{2}

cos (x) = - \frac{2}{2}

cos (x) = - \frac{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = - \frac{2}{2}

General solutions for

cos (x) = - \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

\frac{3 π}{4} + 2 πn :

True

\frac{3 π}{4} + 2 πn

Plug in

n = 1

\frac{3 π}{4} + 2 π 1

For

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

plug in

x = \frac{3 π}{4} + 2 π 1

2 sin (\frac{3 π}{4} + 2 π 1) - 2 cos (\frac{3 π}{4} + 2 π 1) = 2

Refine

2 = 2

\Rightarrow True

Check the solution

\frac{5 π}{4} + 2 πn :

False

\frac{5 π}{4} + 2 πn

Plug in

n = 1

\frac{5 π}{4} + 2 π 1

For

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

plug in

x = \frac{5 π}{4} + 2 π 1

2 sin (\frac{5 π}{4} + 2 π 1) - 2 cos (\frac{5 π}{4} + 2 π 1) = 2

Refine

0 = 2

\Rightarrow False

x = \frac{3 π}{4} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(x)-sqrt(2)cos(x)=2 ?
The general solution for sqrt(2)sin(x)-sqrt(2)cos(x)=2 is x=(3pi)/4+2pin