What is the general solution for 8cos(x)-8sin(x)=0 ?

The general solution for 8cos(x)-8sin(x)=0 is x= pi/4+pin

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8cos(x)-8sin(x)=0

8 cos (x) - 8 sin (x) = 0

x = \frac{π}{4} + πn

Degrees

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

Solution steps

8 cos (x) - 8 sin (x) = 0

Rewrite using trig identities

8 cos (x) - 8 sin (x) = 0

Divide both sides by

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

\frac{8 cos ( x ) - 8 sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

8 - \frac{8 sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

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

8 - 8 tan (x) = 0

8 - 8 tan (x) = 0

Move

8

to the right side

8 - 8 tan (x) = 0

Subtract

8

from both sides

8 - 8 tan (x) - 8 = 0 - 8

Simplify

- 8 tan (x) = - 8

- 8 tan (x) = - 8

Divide both sides by

- 8

- 8 tan (x) = - 8

Divide both sides by

- 8

\frac{- 8 tan ( x )}{- 8} = \frac{- 8}{- 8}

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

sin (4 x) cos (x) - sin (x) cos (4 x) = \frac{1}{2}

arccos (x) = \frac{1}{2}

3 sin^{2} (θ) - 7 sin (θ) + 2 = 0

9 sin^{2} (θ) + 14 cos (θ) - 12 = 5 cos (θ) - 1

4 sec^{2} (x) - 2 = 0

What is the general solution for 8cos(x)-8sin(x)=0 ?
The general solution for 8cos(x)-8sin(x)=0 is x= pi/4+pin