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

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

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2cos(x)tan(x)=1

2 cos (x) tan (x) = 1

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

Rewrite using trig identities

2 cos (x) tan (x)

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)

= 2 sin (x)

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

sin (10 x) = sin (7 x)

185 = \frac{220 2}{π} - cos (π) + cos (x)

x, 2 sin^{2} (x) = sin (x)

3 cos^{2} (θ) - 2 sin (θ) = 1

- 3 cos (2 x) + 3 = 5 sin (x)

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