What is the general solution for -sqrt(1+tan(x))=sec(x) ?

The general solution for -sqrt(1+tan(x))=sec(x) is x=pin,x= pi/4+pin

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-sqrt(1+tan(x))=sec(x)

Solution

- 1 + tan (x) = sec (x)

Solution

x = πn, x = \frac{π}{4} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

- 1 + tan (x) = sec (x)

Square both sides

(- 1 + tan (x))^{2} = sec^{2} (x)

Subtract

sec^{2} (x)

from both sides

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

Rewrite using trig identities

1 - sec^{2} (x) + tan (x)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

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

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

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

Solve by substitution

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

Let:

tan (x) = u

u - u^{2} = 0

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

u - u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 ( - 1 )}, u_{2} = \frac{- 1 - 1}{2 ( - 1 )}

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

\frac{- 1 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 1 = 0

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = 1

tan (x) = 0, tan (x) = 1

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

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 = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

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}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = πn, x = \frac{π}{4} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

- 1 + tan (x) = sec (x)

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

Check the solution

πn :

True

πn

Plug in

n = 1

π 1

For

- 1 + tan (x) = sec (x)

plug in

x = π 1

- 1 + tan (π 1) = sec (π 1)

Refine

- 1 = - 1

\Rightarrow True

Check the solution

\frac{π}{4} + πn :

True

\frac{π}{4} + πn

Plug in

n = 1

\frac{π}{4} + π 1

For

- 1 + tan (x) = sec (x)

plug in

x = \frac{π}{4} + π 1

- 1 + tan (\frac{π}{4} + π 1) = sec (\frac{π}{4} + π 1)

Refine

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow True

x = πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -sqrt(1+tan(x))=sec(x) ?
The general solution for -sqrt(1+tan(x))=sec(x) is x=pin,x= pi/4+pin