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

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

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin (2 x)

from both sides

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

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

Simplify

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

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

Group like terms

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

1 - 1 = 0

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

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

2 cos (x)

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

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

Solving each part separately

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

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

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

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