What is the general solution for ((-sqrt(2))/2)(sin(θ)-cos(θ))=0 ?

The general solution for ((-sqrt(2))/2)(sin(θ)-cos(θ))=0 is θ= pi/4+pin

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((-sqrt(2))/2)(sin(θ)-cos(θ))=0

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

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

\frac{- 2}{2} (sin (θ) - cos (θ)) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

\frac{- tan ( θ ) + 1}{2} = 0

\frac{- tan ( θ ) + 1}{2} = 0

Multiply both sides by

2

\frac{- tan ( θ ) + 1}{2} = 0

Multiply both sides by

2

\frac{( - tan ( θ ) + 1 ) 2}{2} = 0 \cdot 2

Simplify

- tan (θ) + 1 = 0

- tan (θ) + 1 = 0

Move

1

to the right side

- tan (θ) + 1 = 0

Subtract

1

from both sides

- tan (θ) + 1 - 1 = 0 - 1

Simplify

- tan (θ) = - 1

- tan (θ) = - 1

Divide both sides by

- 1

- tan (θ) = - 1

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

3 \cdot sin (x) + 4 \cdot cos (x) = 2

1 = sin (x) + cos (x)

cos (2 t) - cos (t) = - 0.5, 0 < t < 2 π

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

3 sin (x) - 2 cos (x) = 7

What is the general solution for ((-sqrt(2))/2)(sin(θ)-cos(θ))=0 ?
The general solution for ((-sqrt(2))/2)(sin(θ)-cos(θ))=0 is θ= pi/4+pin