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

The general solution for cos(x)(tan(x)csc(x))^2-1=2cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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cos(x)(tan(x)csc(x))^2-1=2cos(x)

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 cos (x)

from both sides

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

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

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

Cancel the common factor:

sin^{2} (x)

= \frac{1 \cdot cos ( x )}{cos ^{2} ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

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

Cancel the common factor:

cos (x)

= \frac{1}{cos ( x )}

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

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}, 2 cos (x) = \frac{2 cos ( x ) cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

- 1 \cdot cos (x) - 2 cos (x) cos (x) + 1

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

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

2 cos (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 3}{- 2 \cdot 2}

Add the numbers:

1 + 3 = 4

= \frac{4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 3}{- 2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = \frac{1}{2}

cos (x) = - 1, cos (x) = \frac{1}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Since the equation is undefined for:

π + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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2 cos (4 θ + 20) = 0.8

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)(tan(x)csc(x))^2-1=2cos(x) ?
The general solution for cos(x)(tan(x)csc(x))^2-1=2cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin