What is the general solution for <= sqrt(3)*cos(x)-sin(x)=0 ?

The general solution for <= sqrt(3)*cos(x)-sin(x)=0 is x= pi/3+pin

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<= sqrt(3)*cos(x)-sin(x)=0

\leq 3 \cdot cos (x) - sin (x) = 0

x = \frac{π}{3} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

\leq 3 cos (x) - sin (x) = 0

Rewrite using trig identities

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

Divide both sides by

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

\frac{3 cos ( x ) - sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

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

3 - tan (x) = 0

3 - tan (x) = 0

Move

3

to the right side

3 - tan (x) = 0

Subtract

3

from both sides

3 - tan (x) - 3 = 0 - 3

Simplify

- tan (x) = - 3

- tan (x) = - 3

Divide both sides by

- 1

- tan (x) = - 3

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 3}{- 1}

Simplify

tan (x) = 3

tan (x) = 3

General solutions for

tan (x) = 3

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}{3} + πn

x = \frac{π}{3} + πn

tan^{2} (x) + sec^{2} (x) = 3

4 sin^{2} (x) - 6 cos^{2} (x) = 0

sin (6 0^{\circ} - a) = 0.598

sin (a) = 0.143

sin (x) = cos^{2} (x) + 1

What is the general solution for <= sqrt(3)*cos(x)-sin(x)=0 ?
The general solution for <= sqrt(3)*cos(x)-sin(x)=0 is x= pi/3+pin