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

The general solution for 2+2cos(x)=2-2sin(x) is x=(3pi)/4+pin

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

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

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2 - 2 sin (x)

from both sides

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

2 + 2 tan (x) = 0

2 + 2 tan (x) = 0

Move

2

to the right side

2 + 2 tan (x) = 0

Subtract

2

from both sides

2 + 2 tan (x) - 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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

sin (x) = 14

tan (x) = 0, 5

2 sec^{2} (x) - 5 sec (x) + 2 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

2 sin (5 x) + 3 = 0, 0 \leq x \leq 2 π

sin (2 θ) = - \frac{24}{25}

What is the general solution for 2+2cos(x)=2-2sin(x) ?
The general solution for 2+2cos(x)=2-2sin(x) is x=(3pi)/4+pin