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

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

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

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

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

- 2 tan (x) + 2 = 0

- 2 tan (x) + 2 = 0

Move

2

to the right side

- 2 tan (x) + 2 = 0

Subtract

2

from both sides

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

Simplify

- 2 tan (x) = - 2

- 2 tan (x) = - 2

Divide both sides by

- 2

- 2 tan (x) = - 2

Divide both sides by

- 2

\frac{- 2 tan ( x )}{- 2} = \frac{- 2}{- 2}

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

cos (3 x) cos (9 x) + sin (3 x) sin (9 x) = 0, (0, 2 π)

1 + cot^{2} (x) + 2 csc (x) = 0

4 sin (2 x) = 3 sin (x)

sin (a) = 0.1245

x, sin (x + z) = cos (y + z)

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