What is the general solution for tan(h)=(sqrt(9))/(9sqrt(3)) ?

The general solution for tan(h)=(sqrt(9))/(9sqrt(3)) is h=0.19012…+pin

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tan(h)=(sqrt(9))/(9sqrt(3))

tan (h) = \frac{9}{9 3}

h = 0.19012 \dots + πn

Solution steps

tan (h) = \frac{9}{9 3}

Simplify

\frac{9}{9 3} : \frac{3}{9}

\frac{9}{9 3}

Factor

9 : 3

Factor

9 = 3^{2}

= 3^{2}

Simplify

3^{2} : 3

3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

= 3

Factor

9 : 3^{2}

Factor

9 = 3^{2}

= \frac{3}{3 ^{2} 3}

Cancel the common factor:

3

= \frac{1}{3 3}

Rationalize

\frac{1}{3 3} : \frac{3}{9}

\frac{1}{3 3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3 3}

1 \cdot 3 = 3

333 = 9

333

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

333 = 3 \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} = 3^{1 + \frac{1}{2} + \frac{1}{2}}

= 3^{1 + \frac{1}{2} + \frac{1}{2}}

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 3^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3^{2}

3^{2} = 9

= 9

= \frac{3}{9}

= \frac{3}{9}

tan (h) = \frac{3}{9}

Apply trig inverse properties

tan (h) = \frac{3}{9}

General solutions for

tan (h) = \frac{3}{9}

tan (x) = a \Rightarrow x = arctan (a) + πn

h = arctan (\frac{3}{9}) + πn

h = arctan (\frac{3}{9}) + πn

Show solutions in decimal form

h = 0.19012 \dots + πn

arcsin (x) + arcsin (3 x) = \frac{π}{2}

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

(cos (x) - 3) (cos (x) + 1) = 0

4 tan^{2} (x) - 16 tan (x) + 7 = 0

1 - cos (x) = 0.5

What is the general solution for tan(h)=(sqrt(9))/(9sqrt(3)) ?
The general solution for tan(h)=(sqrt(9))/(9sqrt(3)) is h=0.19012…+pin