What is the general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) ?

The general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) is

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Popular Trigonometry >

d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x))

Solution

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

Solution

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Solution steps

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )}

Switch sides

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} = d

Subtract

d

from both sides

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d = 0

Simplify

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d : \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d

Convert element to fraction:

d = \frac{d 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

= \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - \frac{d \cdot 4 cos ^{2} ( x )}{4 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 ^{4} ( x ) - cos ^{2} ( x ) + 5 - d \cdot 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )} = 0

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

sin^{4} (x) - cos^{2} (x) + 5 - 4 d cos^{2} (x) = 0

Rewrite using trig identities

5 - cos^{2} (x) + sin^{4} (x) - 4 cos^{2} (x) d

Use the Pythagorean identity:

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

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

= 5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

Simplify

5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d : sin^{4} (x) + sin^{2} (x) + 4 d sin^{2} (x) + 4 - 4 d

5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

= 5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 d (1 - sin^{2} (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)

= 5 - 1 + sin^{2} (x) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 4 d, b = 1, c = sin^{2} (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

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

= 5 - 1 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 d sin^{2} (x)

Subtract the numbers:

5 - 1 = 4

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

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

4 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 sin^{2} (x) d = 0

Solve by substitution

4 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 sin^{2} (x) d = 0

Let:

sin (x) = u

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0 : u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + d = 0

u^{4} + (1 + 4 d) u^{2} + 4 - 4 d = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Solve

v^{2} + (1 + 4 d) v + 4 - 4 d = 0 : v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Solve with the quadratic formula

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Quadratic Equation Formula:

For

a = 1, b = 1 + 4 d, c = 4 - 4 d

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm ( 1 + 4 d ) ^{2} - 4 \cdot 1 \cdot ( 4 - 4 d )}{2 \cdot 1}

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm ( 1 + 4 d ) ^{2} - 4 \cdot 1 \cdot ( 4 - 4 d )}{2 \cdot 1}

Simplify

(1 + 4 d)^{2} - 4 \cdot 1 \cdot (4 - 4 d) : 16 d^{2} + 24 d - 15

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

Multiply the numbers:

4 \cdot 1 = 4

= (4 d + 1)^{2} - 4 (- 4 d + 4)

Expand

(1 + 4 d)^{2} - 4 (4 - 4 d) : 16 d^{2} + 24 d - 15

(1 + 4 d)^{2} - 4 (4 - 4 d)

(1 + 4 d)^{2} : 1 + 8 d + 16 d^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = 4 d

= 1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

Simplify

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2} : 1 + 8 d + 16 d^{2}

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

2 \cdot 1 \cdot 4 d = 8 d

2 \cdot 1 \cdot 4 d

Multiply the numbers:

2 \cdot 1 \cdot 4 = 8

= 8 d

(4 d)^{2} = 16 d^{2}

(4 d)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} d^{2}

4^{2} = 16

= 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2} - 4 (4 - 4 d)

Expand

- 4 (4 - 4 d) : - 16 + 16 d

- 4 (4 - 4 d)

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 4, c = 4 d

= - 4 \cdot 4 - (- 4) \cdot 4 d

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 4 + 4 \cdot 4 d

Multiply the numbers:

4 \cdot 4 = 16

= - 16 + 16 d

= 1 + 8 d + 16 d^{2} - 16 + 16 d

Simplify

1 + 8 d + 16 d^{2} - 16 + 16 d : 16 d^{2} + 24 d - 15

1 + 8 d + 16 d^{2} - 16 + 16 d

Group like terms

= 16 d^{2} + 8 d + 16 d + 1 - 16

Add similar elements:

8 d + 16 d = 24 d

= 16 d^{2} + 24 d + 1 - 16

Add/Subtract the numbers:

1 - 16 = - 15

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}, v_{2} = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

v = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- ( 4 d + 1 ) + 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

Distribute parentheses

= - (1) - (4 d)

Apply minus-plus rules

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

v = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- ( 4 d + 1 ) - 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

Distribute parentheses

= - (1) - (4 d)

Apply minus-plus rules

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

The solutions to the quadratic equation are:

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Solve

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

The solutions are

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Substitute back

u = sin (x)

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

General solutions for

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

General solutions for

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

General solutions for

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

General solutions for

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Combine all the solutions

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) ?
The general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) is