What is the general solution for 2cos^2(x)-2sin^2(x)=0 ?

The general solution for 2cos^2(x)-2sin^2(x)=0 is x=(3pi)/4+pin,x= pi/4+pin

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Popular >

2cos^2(x)-2sin^2(x)=0

Solution

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

Solution

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Factor

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

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

Factor out common term

2

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) = 0

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

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

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Popular Examples

csc^2(θ)+5=7-csc(θ)

csc^{2} (θ) + 5 = 7 - csc (θ)

3tan^2(x/3)-1=0

3 tan^{2} (\frac{x}{3}) - 1 = 0

cos(x)= 7/10

cos (x) = \frac{7}{10}

2cos(2x)+3=2

2 cos (2 x) + 3 = 2

2cos(x)=sec(x)

2 cos (x) = sec (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-2sin^2(x)=0 ?
The general solution for 2cos^2(x)-2sin^2(x)=0 is x=(3pi)/4+pin,x= pi/4+pin