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

The general solution for 2cos^2(2x)+sin(2x)=1 is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin,x= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{π}{4} + πn

Degrees

x = 10 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

sin (2 x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 1

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

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 x)

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

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

sin (2 x) = - \frac{1}{2} : x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

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

General solutions for

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

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{7 π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

2 x = \frac{7 π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

Solve

2 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{12} + πn

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{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{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

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

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

Solve

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{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{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

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{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{π}{4} + πn

Popular Examples

sin(4θ)+sin(2θ)=0

sin (4 θ) + sin (2 θ) = 0

sec(x)= 2/(sqrt(3))

sec (x) = \frac{2}{3}

2sin(x)+1=3csc(x)

2 sin (x) + 1 = 3 csc (x)

14sin(20x)-3=2

14 sin (20 x) - 3 = 2

sqrt(3)tan(3x)-1=0

3 tan (3 x) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(2x)+sin(2x)=1 ?
The general solution for 2cos^2(2x)+sin(2x)=1 is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin,x= pi/4+pin