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

The general solution for tan^2(x)+cos^2(x)-1=0 is x=2pin,x=pi+2pin

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

Apply exponent rule:

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

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 1 + 1 - sin^{2} (x) + \frac{sin ^{2} ( x )}{1 - sin ^{2} ( x )}

Simplify

- 1 + 1 - sin^{2} (x) + \frac{sin ^{2} ( x )}{1 - sin ^{2} ( x )} : \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

- 1 + 1 - sin^{2} (x) + \frac{sin ^{2} ( x )}{1 - sin ^{2} ( x )}

- 1 + 1 = 0

= \frac{sin ^{2} ( x )}{- sin ^{2} ( x ) + 1} - sin^{2} (x)

Convert element to fraction:

sin^{2} (x) = \frac{sin ^{2} ( x ) ( - sin ^{2} ( x ) + 1 )}{- sin ^{2} ( x ) + 1}

= \frac{sin ^{2} ( x )}{- sin ^{2} ( x ) + 1} - \frac{sin ^{2} ( x ) ( - sin ^{2} ( x ) + 1 )}{- sin ^{2} ( x ) + 1}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( x ) - sin ^{2} ( x ) ( - sin ^{2} ( x ) + 1 )}{- sin ^{2} ( x ) + 1}

Expand

sin^{2} (x) - sin^{2} (x) (- sin^{2} (x) + 1) : sin^{4} (x)

sin^{2} (x) - sin^{2} (x) (- sin^{2} (x) + 1)

Expand

- sin^{2} (x) (- sin^{2} (x) + 1) : sin^{4} (x) - sin^{2} (x)

- sin^{2} (x) (- sin^{2} (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = - sin^{2} (x), b = - sin^{2} (x), c = 1

= - sin^{2} (x) (- sin^{2} (x)) + (- sin^{2} (x)) \cdot 1

Apply minus-plus rules

- (- a) = a, + (- a) = - a

= sin^{2} (x) sin^{2} (x) - 1 \cdot sin^{2} (x)

Simplify

sin^{2} (x) sin^{2} (x) - 1 \cdot sin^{2} (x) : sin^{4} (x) - sin^{2} (x)

sin^{2} (x) sin^{2} (x) - 1 \cdot sin^{2} (x)

sin^{2} (x) sin^{2} (x) = sin^{4} (x)

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

Apply exponent rule:

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

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

= sin^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= sin^{4} (x)

1 \cdot sin^{2} (x) = sin^{2} (x)

1 \cdot sin^{2} (x)

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

= sin^{2} (x)

= sin^{4} (x) - sin^{2} (x)

= sin^{4} (x) - sin^{2} (x)

= sin^{2} (x) + sin^{4} (x) - sin^{2} (x)

Add similar elements:

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

= sin^{4} (x)

= \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

= \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

\frac{sin ^{4} ( x )}{1 - sin ^{2} ( x )} = 0

Solve by substitution

\frac{sin ^{4} ( x )}{1 - sin ^{2} ( x )} = 0

Let:

sin (x) = u

\frac{u ^{4}}{1 - u ^{2}} = 0

\frac{u ^{4}}{1 - u ^{2}} = 0 : u = 0

\frac{u ^{4}}{1 - u ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

u^{4} = 0

Solve

u^{4} = 0 : u = 0

u^{4} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0

The solution is

u = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

\frac{u ^{4}}{1 - u ^{2}}

and compare to zero

Solve

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

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

u = 0

Substitute back

u = sin (x)

sin (x) = 0

sin (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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sin(2a+10)=cos(3a-20)

sin (2 a + 10) = cos (3 a - 20)

b= 3/((cot(x)))

b = \frac{3}{( cot ( x ) )}

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Frequently Asked Questions (FAQ)

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