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

The general solution for 2cos(2x)-1=1-2sin(2x) is x=2pin,x=pi+2pin,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(2x)-1=1-2sin(2x)

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1 - 2 sin (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

= - 2 + 2 cos (2 x) + 4 sin (x) cos (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

- 2 + 2 = 0

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

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

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

Factor

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

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

Apply exponent rule:

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

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

= - 4 sin (x) sin (x) + 4 sin (x) cos (x)

Factor out common term

4 sin (x)

= 4 sin (x) (- sin (x) + cos (x))

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

- sin (x) + cos (x) = 0 : x = \frac{π}{4} + πn

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

Rewrite using trig identities

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

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{- sin ( x ) + cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

- \frac{sin ( x )}{cos ( x )} + 1 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- tan (x) + 1 = 0

- tan (x) + 1 = 0

Move

1

to the right side

- tan (x) + 1 = 0

Subtract

1

from both sides

- tan (x) + 1 - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

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

Popular Examples

cot^2(θ)+3csc(θ)=-3

cot^{2} (θ) + 3 csc (θ) = - 3

3tan^2(α)=3tan(α)

3 tan^{2} (α) = 3 tan (α)

cos(4x)+cos(2x)=0

cos (4 x) + cos (2 x) = 0

sin(x)= 10/50

sin (x) = \frac{10}{50}

6sin(4x)+4=5,0<= x<= 2pi

6 sin (4 x) + 4 = 5, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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