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

The general solution for cos(x)tan(x)-2cos^2(x)=-1 is x= pi/6+2pin,x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- 1

from both sides

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= sin (x)

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

1 - 2 = - 1

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 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 \cdot 2}

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Subtract 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

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

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 = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

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2sin(pi/4-x)+sqrt(3)=0

2 sin (\frac{π}{4} - x) + 3 = 0

3.1=48.1(1-cos((28.66x)/(48.1)))

3.1 = 48.1 (1 - cos (\frac{28.66 x}{48.1}))

sin^2(x)-2cos(x)=-2

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

2cos^2(x)+6=7

2 cos^{2} (x) + 6 = 7

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)tan(x)-2cos^2(x)=-1 ?
The general solution for cos(x)tan(x)-2cos^2(x)=-1 is x= pi/6+2pin,x=(5pi)/6+2pin