What is the general solution for 2cos(x)*tan(x)-1=0 ?

The general solution for 2cos(x)*tan(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2cos(x)*tan(x)-1=0

2 cos (x) \cdot tan (x) - 1 = 0

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) tan (x) - 1 = 0

Rewrite using trig identities

2 cos (x) tan (x) - 1

cos (x) tan (x) = sin (x)

cos (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

cos (x) \frac{sin ( x )}{cos ( x )} : sin (x)

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

cos (x)

= sin (x)

= sin (x)

= - 1 + 2 sin (x)

- 1 + 2 sin (x) = 0

Move

1

to the right side

- 1 + 2 sin (x) = 0

Add

1

to both sides

- 1 + 2 sin (x) + 1 = 0 + 1

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

What is the general solution for 2cos(x)*tan(x)-1=0 ?
The general solution for 2cos(x)*tan(x)-1=0 is x= pi/6+2pin,x=(5pi)/6+2pin