What is the general solution for sin^2(x)=1-tan^2(x),0<= x<= pi/2 ?

The general solution for sin^2(x)=1-tan^2(x),0<= x<= pi/2 is x=0.66623…

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sin^2(x)=1-tan^2(x),0<= x<= pi/2

Solution

sin^{2} (x) = 1 - tan^{2} (x), 0 \leq x \leq \frac{π}{2}

Solution

x = 0.66623 \dots

Degrees

x = 38.17270 \dots^{\circ}

Solution steps

sin^{2} (x) = 1 - tan^{2} (x), 0 \leq x \leq \frac{π}{2}

Subtract

1 - tan^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (tan (x) + cos (x)) (tan (x) - cos (x))

(tan (x) + cos (x)) (tan (x) - cos (x)) = 0

Solving each part separately

tan (x) + cos (x) = 0 or tan (x) - cos (x) = 0

tan (x) + cos (x) = 0, 0 \leq x \leq \frac{π}{2} :

No Solution

tan (x) + cos (x) = 0, 0 \leq x \leq \frac{π}{2}

Express with sin, cos

cos (x) + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (x) cos (x) + sin (x)

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2} :

No Solution

sin (x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 5}{2}

General solutions for

sin (x) = - \frac{- 1 + 5}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Solutions for the range

0 \leq x \leq \frac{π}{2}

No Solution

sin (x) = \frac{1 + 5}{2}, 0 \leq x \leq \frac{π}{2} :

No Solution

sin (x) = \frac{1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution

tan (x) - cos (x) = 0, 0 \leq x \leq \frac{π}{2} : x = arcsin (\frac{5 - 1}{2})

tan (x) - cos (x) = 0, 0 \leq x \leq \frac{π}{2}

Express with sin, cos

- cos (x) + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

- cos (x) + \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (x) cos (x) + sin (x)

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = sin (x)

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2} : x = arcsin (\frac{5 - 1}{2})

sin (x) = \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 + 5}{2}

General solutions for

sin (x) = \frac{- 1 + 5}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Solutions for the range

0 \leq x \leq \frac{π}{2}

x = arcsin (\frac{5 - 1}{2})

sin (x) = \frac{- 1 - 5}{2}, 0 \leq x \leq \frac{π}{2} :

No Solution

sin (x) = \frac{- 1 - 5}{2}, 0 \leq x \leq \frac{π}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{5 - 1}{2})

Combine all the solutions

x = arcsin (\frac{5 - 1}{2})

Show solutions in decimal form

x = 0.66623 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)=1-tan^2(x),0<= x<= pi/2 ?
The general solution for sin^2(x)=1-tan^2(x),0<= x<= pi/2 is x=0.66623…