What is the general solution for sqrt(cos^2(2x)+1)+sin(2x)=2 ?

The general solution for sqrt(cos^2(2x)+1)+sin(2x)=2 is x= pi/4+pin

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Popular Trigonometry >

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

Solution

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

Solution

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (2 x) = u

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

- 2 + u + 2 - u^{2} = 0 : u = 1

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

Remove square roots

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

Subtract

u

from both sides

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

Simplify

2 - u^{2} - 2 = - u

Add

2

to both sides

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

Simplify

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

Square both sides:

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

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

(2 - u^{2})^{2} = (- u + 2)^{2}

Expand

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

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

Apply radical rule:

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

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

Apply exponent rule:

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

= (2 - u^{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 - u^{2}

Expand

(- u + 2)^{2} : u^{2} - 4 u + 4

(- u + 2)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - u, b = 2

= (- u)^{2} + 2 (- u) \cdot 2 + 2^{2}

Simplify

(- u)^{2} + 2 (- u) \cdot 2 + 2^{2} : u^{2} - 4 u + 4

(- u)^{2} + 2 (- u) \cdot 2 + 2^{2}

Remove parentheses:

(- a) = - a

= (- u)^{2} - 2 u \cdot 2 + 2^{2}

Apply exponent rule:

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

n

is even

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

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

Refine

= u^{2} - 4 u + 4

= u^{2} - 4 u + 4

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

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

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

Solve

2 - u^{2} = u^{2} - 4 u + 4 : u = 1

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

Switch sides

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

Move

u^{2}

to the left side

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

Add

u^{2}

to both sides

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

Simplify

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

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

Move

2

to the left side

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

Subtract

2

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

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

\frac{- ( - 4 )}{2 \cdot 2} = 1

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

u = 1

Verify Solutions:

u = 1

True

Check the solutions by plugging them into

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

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

Plug in

u = 1 :

True

- 2 + 1 + 2 - 1^{2} = 0

- 2 + 1 + 2 - 1^{2} = 0

- 2 + 1 + 2 - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= - 2 + 1 + 2 - 1

2 - 1 = 1

2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

Apply rule

1 = 1

= 1

= - 2 + 1 + 1

Add/Subtract the numbers:

- 2 + 1 + 1 = 0

= 0

0 = 0

True

The solution is

u = 1

Substitute back

u = sin (2 x)

sin (2 x) = 1

sin (2 x) = 1

sin (2 x) = 1 : x = \frac{π}{4} + πn

sin (2 x) = 1

General solutions for

sin (2 x) = 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}{2} + 2 πn

2 x = \frac{π}{2} + 2 πn

Solve

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

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn

Graph

Popular Examples

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(cos^2(2x)+1)+sin(2x)=2 ?
The general solution for sqrt(cos^2(2x)+1)+sin(2x)=2 is x= pi/4+pin