What is the general solution for 9cos(x)-9sin(x)=0 ?

The general solution for 9cos(x)-9sin(x)=0 is x= pi/4+pin

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9cos(x)-9sin(x)=0

9 cos (x) - 9 sin (x) = 0

x = \frac{π}{4} + πn

Degrees

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

Solution steps

9 cos (x) - 9 sin (x) = 0

Rewrite using trig identities

9 cos (x) - 9 sin (x) = 0

Divide both sides by

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

\frac{9 cos ( x ) - 9 sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

9 - \frac{9 sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

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

9 - 9 tan (x) = 0

9 - 9 tan (x) = 0

Move

9

to the right side

9 - 9 tan (x) = 0

Subtract

9

from both sides

9 - 9 tan (x) - 9 = 0 - 9

Simplify

- 9 tan (x) = - 9

- 9 tan (x) = - 9

Divide both sides by

- 9

- 9 tan (x) = - 9

Divide both sides by

- 9

\frac{- 9 tan ( x )}{- 9} = \frac{- 9}{- 9}

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^{2} (3 x) - sin^{2} (x) = 0

sec^{2} (x) + 6 tan (x) + 4 = 0

1 - 2 tan^{2} (θ) = - tan^{2} (θ)

cos (a) = \frac{1}{3}

2 cos (2 θ) = 3

What is the general solution for 9cos(x)-9sin(x)=0 ?
The general solution for 9cos(x)-9sin(x)=0 is x= pi/4+pin