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

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

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

Solution

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

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

Subtract

2 cos^{2} (x)

from both sides

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

Factor

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

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

Factor out common term

2

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

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

sin (x) + cos (x) = 0

Rewrite using trig identities

sin (x) + cos (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 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

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

sin (x) - cos (x) = 0

Rewrite using trig identities

sin (x) - cos (x) = 0

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) - 1 = 0

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 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{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

cos^2(x)=((6k-7))/7

cos^{2} (x) = \frac{( 6 k - 7 )}{7}

((1+tanh(x))/(1-tanh(x)))^2=2

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

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

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

cos(3x)=sin(x)

cos (3 x) = sin (x)

4tan^2(θ)=12,0<= θ<2pi

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

Frequently Asked Questions (FAQ)

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