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

The general solution for sin(2x)=-cos(2x) is x=(3pi)/8+(pin)/2

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

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

x = \frac{3 π}{8} + \frac{πn}{2}

Degrees

x = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

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{3 π}{4} + πn

2 x = \frac{3 π}{4} + πn

Solve

2 x = \frac{3 π}{4} + πn : x = \frac{3 π}{8} + \frac{πn}{2}

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{3 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{3 π}{4}}{2} + \frac{πn}{2} : \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

3 sin (4 x) + 5 = 6

4 sin (x) + 3 = 0

tan (2 θ) = 3

cos (x) = 0.2

sin (2 x) - \frac{3}{2} = 0

What is the general solution for sin(2x)=-cos(2x) ?
The general solution for sin(2x)=-cos(2x) is x=(3pi)/8+(pin)/2