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

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

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sin^2(x)+tan^2(x)=0

Solution

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

sin^{2} (x) + tan^{2} (x) = 0

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

sin^{2} (x) + (\frac{sin ( x )}{cos ( x )})^{2} : \frac{sin ^{2} ( x ) cos ^{2} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

\frac{sin ^{2} ( x ) + cos ^{2} ( x ) sin ^{2} ( x )}{cos ^{2} ( x )} = 0

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

sin^{2} (x) + cos^{2} (x) sin^{2} (x) = 0

Factor

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

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

Factor out common term

sin^{2} (x)

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

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

Solving each part separately

sin^{2} (x) = 0 or cos^{2} (x) + 1 = 0

sin^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

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

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

No Solution

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} + 1 = 0

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Substitute back

u = cos (x)

cos (x) = i, cos (x) = - i

cos (x) = i, cos (x) = - i

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

tan(x)= 9/12

tan (x) = \frac{9}{12}

cos(x)= 24/25

cos (x) = \frac{24}{25}

2(tan(θ)+3)=5+tan(θ),0<= θ<2pi

2 (tan (θ) + 3) = 5 + tan (θ), 0 \leq θ < 2 π

3cos(3θ)=2cos(θ)

3 cos (3 θ) = 2 cos (θ)

(1-csc(θ))/(cot(θ))=tan(θ)-csc(θ)

\frac{1 - csc ( θ )}{cot ( θ )} = tan (θ) - csc (θ)

Frequently Asked Questions (FAQ)

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