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

The general solution for tan(x)-sqrt(1-2tan^2(x))=0 is x=0.52359…+pin

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Popular Trigonometry >

tan(x)-sqrt(1-2tan^2(x))=0

Solution

tan (x) - 1 - 2 tan^{2} (x) = 0

Solution

x = 0.52359 \dots + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (x) - 1 - 2 tan^{2} (x) = 0

Solve by substitution

tan (x) - 1 - 2 tan^{2} (x) = 0

Let:

tan (x) = u

u - 1 - 2 u^{2} = 0

u - 1 - 2 u^{2} = 0 : u = \frac{1}{3}

u - 1 - 2 u^{2} = 0

Remove square roots

u - 1 - 2 u^{2} = 0

Subtract

u

from both sides

u - 1 - 2 u^{2} - u = 0 - u

Simplify

- 1 - 2 u^{2} = - u

Square both sides:

1 - 2 u^{2} = u^{2}

u - 1 - 2 u^{2} = 0

(- 1 - 2 u^{2})^{2} = (- u)^{2}

Expand

(- 1 - 2 u^{2})^{2} : 1 - 2 u^{2}

(- 1 - 2 u^{2})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1 - 2 u^{2})^{2} = (1 - 2 u^{2})^{2}

= (1 - 2 u^{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((1 - 2 u^{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (1 - 2 u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - 2 u^{2}

Expand

(- u)^{2} : u^{2}

(- u)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- u)^{2} = u^{2}

= u^{2}

1 - 2 u^{2} = u^{2}

1 - 2 u^{2} = u^{2}

1 - 2 u^{2} = u^{2}

Solve

1 - 2 u^{2} = u^{2} : u = \frac{1}{3}, u = - \frac{1}{3}

1 - 2 u^{2} = u^{2}

Move

1

to the right side

1 - 2 u^{2} = u^{2}

Subtract

1

from both sides

1 - 2 u^{2} - 1 = u^{2} - 1

Simplify

- 2 u^{2} = u^{2} - 1

- 2 u^{2} = u^{2} - 1

Move

u^{2}

to the left side

- 2 u^{2} = u^{2} - 1

Subtract

u^{2}

from both sides

- 2 u^{2} - u^{2} = u^{2} - 1 - u^{2}

Simplify

- 3 u^{2} = - 1

- 3 u^{2} = - 1

Divide both sides by

- 3

- 3 u^{2} = - 1

Divide both sides by

- 3

\frac{- 3 u ^{2}}{- 3} = \frac{- 1}{- 3}

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

Verify Solutions:

u = \frac{1}{3}

True

, u = - \frac{1}{3}

False

Check the solutions by plugging them into

u - 1 - 2 u^{2} = 0

Remove the ones that don't agree with the equation.

Plug in

u = \frac{1}{3} :

True

\frac{1}{3} - 1 - 2 (\frac{1}{3})^{2} = 0

\frac{1}{3} - 1 - 2 (\frac{1}{3})^{2} = 0

\frac{1}{3} - 1 - 2 (\frac{1}{3})^{2}

1 - 2 (\frac{1}{3})^{2} = \frac{1}{3}

1 - 2 (\frac{1}{3})^{2}

2 (\frac{1}{3})^{2} = \frac{2}{3}

2 (\frac{1}{3})^{2}

(\frac{1}{3})^{2} = \frac{1}{3}

(\frac{1}{3})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((\frac{1}{3})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (\frac{1}{3})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= \frac{1}{3}

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{3}

= 1 - \frac{2}{3}

Join

1 - \frac{2}{3} : \frac{1}{3}

1 - \frac{2}{3}

Convert element to fraction:

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

= \frac{1 \cdot 3}{3} - \frac{2}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 3 - 2}{3}

1 \cdot 3 - 2 = 1

1 \cdot 3 - 2

Multiply the numbers:

1 \cdot 3 = 3

= 3 - 2

Subtract the numbers:

3 - 2 = 1

= 1

= \frac{1}{3}

= \frac{1}{3}

= \frac{1}{3} - \frac{1}{3}

Add similar elements:

\frac{1}{3} - \frac{1}{3} = 0

= 0

0 = 0

True

Plug in

u = - \frac{1}{3} :

False

(- \frac{1}{3}) - 1 - 2 (- \frac{1}{3})^{2} = 0

(- \frac{1}{3}) - 1 - 2 (- \frac{1}{3})^{2} = - 2 \frac{1}{3}

(- \frac{1}{3}) - 1 - 2 (- \frac{1}{3})^{2}

Remove parentheses:

(- a) = - a

= - \frac{1}{3} - 1 - 2 (- \frac{1}{3})^{2}

1 - 2 (- \frac{1}{3})^{2} = \frac{1}{3}

1 - 2 (- \frac{1}{3})^{2}

2 (- \frac{1}{3})^{2} = \frac{2}{3}

2 (- \frac{1}{3})^{2}

(- \frac{1}{3})^{2} = \frac{1}{3}

(- \frac{1}{3})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{1}{3})^{2} = (\frac{1}{3})^{2}

= (\frac{1}{3})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((\frac{1}{3})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (\frac{1}{3})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= \frac{1}{3}

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{3}

= 1 - \frac{2}{3}

Join

1 - \frac{2}{3} : \frac{1}{3}

1 - \frac{2}{3}

Convert element to fraction:

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

= \frac{1 \cdot 3}{3} - \frac{2}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 3 - 2}{3}

1 \cdot 3 - 2 = 1

1 \cdot 3 - 2

Multiply the numbers:

1 \cdot 3 = 3

= 3 - 2

Subtract the numbers:

3 - 2 = 1

= 1

= \frac{1}{3}

= \frac{1}{3}

= - \frac{1}{3} - \frac{1}{3}

Add similar elements:

- \frac{1}{3} - \frac{1}{3} = - 2 \frac{1}{3}

= - 2 \frac{1}{3}

- 2 \frac{1}{3} = 0

False

The solution is

u = \frac{1}{3}

Substitute back

u = tan (x)

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

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

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn

Show solutions in decimal form

x = 0.52359 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)-sqrt(1-2tan^2(x))=0 ?
The general solution for tan(x)-sqrt(1-2tan^2(x))=0 is x=0.52359…+pin