What is the general solution for cos(2θ)=sin(2θ) ?

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

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cos(2θ)=sin(2θ)

cos (2 θ) = sin (2 θ)

θ = \frac{π}{8} + \frac{πn}{2}

Degrees

θ = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

cos (2 θ) = sin (2 θ)

Rewrite using trig identities

cos (2 θ) = sin (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{cos ( 2 θ )}{cos ( 2 θ )} = \frac{sin ( 2 θ )}{cos ( 2 θ )}

Simplify

1 = \frac{sin ( 2 θ )}{cos ( 2 θ )}

Use the basic trigonometric identity:

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

1 = tan (2 θ)

Switch sides

tan (2 θ) = 1

tan (2 θ) = 1

General solutions for

tan (2 θ) = 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 θ = \frac{π}{4} + πn

2 θ = \frac{π}{4} + πn

Solve

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

2 θ = \frac{π}{4} + πn

Divide both sides by

2

2 θ = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

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

tan (x) (cot (x) - \frac{3}{3}) = 0

24 sin^{2} (x) = 24 - 12 cos (x)

3 tan (x) = 1

2 cos^{2} (t) + 3 cos (t) + 1 = 0

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