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

The general solution for cos(2x)=sin(2x)+1 is x=2pin,x=pi+2pin,x=(3pi)/4+pin

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin (2 x) + 1

from both sides

cos (2 x) - sin (2 x) - 1 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

- 2 sin (x)

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

- 2 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{3 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

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

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

Popular Examples

cos(x)= 4/5

cos (x) = \frac{4}{5}

tan(x)-1=0

tan (x) - 1 = 0

tan^2(θ)-3tan(θ)+2=0

tan^{2} (θ) - 3 tan (θ) + 2 = 0

tan(θ)= 1/(sqrt(3))

tan (θ) = \frac{1}{3}

2cos^2(x)-cos(x)=0

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

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=sin(2x)+1 ?
The general solution for cos(2x)=sin(2x)+1 is x=2pin,x=pi+2pin,x=(3pi)/4+pin