What is the general solution for 4cos^2(x)+3sin^2(x)-4=-tan^2(x) ?

The general solution for 4cos^2(x)+3sin^2(x)-4=-tan^2(x) is x=2pin,x=pi+2pin

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4cos^2(x)+3sin^2(x)-4=-tan^2(x)

Solution

4 cos^{2} (x) + 3 sin^{2} (x) - 4 = - tan^{2} (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) + 3 sin^{2} (x) - 4 = - tan^{2} (x)

Subtract

- tan^{2} (x)

from both sides

4 cos^{2} (x) + 3 sin^{2} (x) - 4 + tan^{2} (x) = 0

Rewrite using trig identities

- 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 (1 - sin^{2} (x))

Simplify

- 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 (1 - sin^{2} (x)) : tan^{2} (x) - sin^{2} (x)

- 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 (1 - sin^{2} (x))

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

= - 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 - 4 sin^{2} (x)

Simplify

- 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 - 4 sin^{2} (x) : tan^{2} (x) - sin^{2} (x)

- 4 + tan^{2} (x) + 3 sin^{2} (x) + 4 - 4 sin^{2} (x)

Group like terms

= tan^{2} (x) + 3 sin^{2} (x) - 4 sin^{2} (x) - 4 + 4

Add similar elements:

3 sin^{2} (x) - 4 sin^{2} (x) = - sin^{2} (x)

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

- 4 + 4 = 0

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

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

= (tan (x) + sin (x)) (tan (x) - sin (x))

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

Solving each part separately

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

tan (x) + sin (x) = 0 : x = 2 πn, x = π + 2 πn

tan (x) + sin (x) = 0

Express with sin, cos

sin (x) + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

sin (x) + cos (x) sin (x) = 0

Factor

sin (x) + cos (x) sin (x) : sin (x) (cos (x) + 1)

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

Factor out common term

sin (x)

= sin (x) (1 + cos (x))

sin (x) (cos (x) + 1) = 0

Solving each part separately

sin (x) = 0 or cos (x) + 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

cos (x) + 1 - 1 = 0 - 1

Simplify

cos (x) = - 1

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

Combine all the solutions

x = 2 πn, x = π + 2 πn

tan (x) - sin (x) = 0 : x = 2 πn, x = π + 2 πn

tan (x) - sin (x) = 0

Express with sin, cos

- sin (x) + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

sin (x) - cos (x) sin (x) = 0

Factor

sin (x) - cos (x) sin (x) : - sin (x) (cos (x) - 1)

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

Factor out common term

- sin (x)

= - sin (x) (- 1 + cos (x))

- sin (x) (cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or cos (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

cos (x) - 1 + 1 = 0 + 1

Simplify

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+3sin^2(x)-4=-tan^2(x) ?
The general solution for 4cos^2(x)+3sin^2(x)-4=-tan^2(x) is x=2pin,x=pi+2pin