What is the general solution for sin(2x)-cos(2x)=0 ?

The general solution for sin(2x)-cos(2x)=0 is x= pi/8+(pin)/2

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

Solution

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

Solution

x = \frac{π}{8} + \frac{πn}{2}

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

\frac{sin ( 2 x ) - cos ( 2 x )}{cos ( 2 x )} = \frac{0}{cos ( 2 x )}

Simplify

\frac{sin ( 2 x )}{cos ( 2 x )} - 1 = 0

Use the basic trigonometric identity:

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

tan (2 x) - 1 = 0

tan (2 x) - 1 = 0

Move

1

to the right side

tan (2 x) - 1 = 0

Add

1

to both sides

tan (2 x) - 1 + 1 = 0 + 1

Simplify

tan (2 x) = 1

tan (2 x) = 1

General solutions for

tan (2 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}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

2 x = \frac{π}{4} + πn : x = \frac{π}{8} + \frac{πn}{2}

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

Popular Examples

tan(x)-3cot(x)=0

tan (x) - 3 cot (x) = 0

sin(θ/2)=-1/2

sin (\frac{θ}{2}) = - \frac{1}{2}

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

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

cos(x+(3pi)/4)-cos(x-(3pi)/4)=1

cos (x + \frac{3 π}{4}) - cos (x - \frac{3 π}{4}) = 1

tan(2θ)+2cos(θ)=0

tan (2 θ) + 2 cos (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(2x)-cos(2x)=0 ?
The general solution for sin(2x)-cos(2x)=0 is x= pi/8+(pin)/2