What is the general solution for 4cos(x)-2=2tan(x)cot(x)-sec(x) ?

The general solution for 4cos(x)-2=2tan(x)cot(x)-sec(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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4cos(x)-2=2tan(x)cot(x)-sec(x)

Solution

4 cos (x) - 2 = 2 tan (x) cot (x) - sec (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

4 cos (x) - 2 = 2 tan (x) cot (x) - sec (x)

Subtract

2 tan (x) cot (x) - sec (x)

from both sides

4 cos (x) - 2 - 2 tan (x) cot (x) + sec (x) = 0

Express with sin, cos

- 2 + sec (x) + 4 cos (x) - 2 cot (x) tan (x)

Use the basic trigonometric identity:

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

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 cot (x) tan (x)

Use the basic trigonometric identity:

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

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

- 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= \frac{sin ( x ) \cdot 2}{sin ( x )}

Cancel the common factor:

sin (x)

= 2

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

Group like terms

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

Subtract the numbers:

- 2 - 2 = - 4

= \frac{1}{cos ( x )} + 4 cos (x) - 4

Convert element to fraction:

4 cos (x) = \frac{4 cos ( x ) cos ( x )}{cos ( x )}, 4 = \frac{4 cos ( x )}{cos ( x )}

= \frac{1}{cos ( x )} + \frac{4 cos ( x ) cos ( x )}{cos ( x )} - \frac{4 cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + 4 cos ( x ) cos ( x ) - 4 cos ( x )}{cos ( x )}

1 + 4 cos (x) cos (x) - 4 cos (x) = 1 + 4 cos^{2} (x) - 4 cos (x)

1 + 4 cos (x) cos (x) - 4 cos (x)

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

4 cos (x) cos (x)

Apply exponent rule:

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

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

= 4 cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 4 cos^{2} (x)

= 1 + 4 cos^{2} (x) - 4 cos (x)

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

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

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

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

1 - 4 cos (x) + 4 cos^{2} (x) = 0

Solve by substitution

1 - 4 cos (x) + 4 cos^{2} (x) = 0

Let:

cos (x) = u

1 - 4 u + 4 u^{2} = 0

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

1 - 4 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

4 u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

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

Apply exponent rule:

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

n

is even

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

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

\frac{- ( - 4 )}{2 \cdot 4}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (x)

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

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

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

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(x)-2=2tan(x)cot(x)-sec(x) ?
The general solution for 4cos(x)-2=2tan(x)cot(x)-sec(x) is x= pi/3+2pin,x=(5pi)/3+2pin