What is the general solution for 1+cos^2(x^2+1)=sin^2(x^2+1) ?

The general solution for 1+cos^2(x^2+1)=sin^2(x^2+1) is x=sqrt((pi+2pin-2)/2),x=-sqrt((pi+2pin-2)/2)

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

Solution

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

Solution

x = \frac{π + 2 πn - 2}{2}, x = - \frac{π + 2 πn - 2}{2}

Degrees

x = 0^{\circ} + 110.39499 \dots^{\circ} n, x = 0^{\circ} - 110.39499 \dots^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

General solutions for

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

2 (1 + x^{2}) = π + 2 πn

2 (1 + x^{2}) = π + 2 πn

Solve

2 (1 + x^{2}) = π + 2 πn : x = \frac{π + 2 πn - 2}{2}, x = - \frac{π + 2 πn - 2}{2}

2 (1 + x^{2}) = π + 2 πn

Divide both sides by

2

2 (1 + x^{2}) = π + 2 πn

Divide both sides by

2

\frac{2 ( 1 + x ^{2} )}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

1 + x^{2} = \frac{π}{2} + πn

1 + x^{2} = \frac{π}{2} + πn

Move

1

to the right side

1 + x^{2} = \frac{π}{2} + πn

Subtract

1

from both sides

1 + x^{2} - 1 = \frac{π}{2} + πn - 1

Simplify

x^{2} = \frac{π}{2} + πn - 1

x^{2} = \frac{π}{2} + πn - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{π}{2} + πn - 1, x = - \frac{π}{2} + πn - 1

Simplify

\frac{π}{2} + πn - 1 : \frac{π + 2 πn - 2}{2}

\frac{π}{2} + πn - 1

Join

\frac{π}{2} + πn - 1 : \frac{π + 2 πn - 2}{2}

\frac{π}{2} + πn - 1

Convert element to fraction:

πn = \frac{πn 2}{2}, 1 = \frac{1 \cdot 2}{2}

= \frac{π}{2} + \frac{πn \cdot 2}{2} - \frac{1 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + πn \cdot 2 - 1 \cdot 2}{2}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{π + 2 πn - 2}{2}

= \frac{π + πn \cdot 2 - 2}{2}

= \frac{π + 2 πn - 2}{2}

Simplify

- \frac{π}{2} + πn - 1 : - \frac{π + 2 πn - 2}{2}

- \frac{π}{2} + πn - 1

Join

\frac{π}{2} + πn - 1 : \frac{π + 2 πn - 2}{2}

\frac{π}{2} + πn - 1

Convert element to fraction:

πn = \frac{πn 2}{2}, 1 = \frac{1 \cdot 2}{2}

= \frac{π}{2} + \frac{πn \cdot 2}{2} - \frac{1 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + πn \cdot 2 - 1 \cdot 2}{2}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{π + 2 πn - 2}{2}

= - \frac{π + 2 πn - 2}{2}

x = \frac{π + 2 πn - 2}{2}, x = - \frac{π + 2 πn - 2}{2}

x = \frac{π + 2 πn - 2}{2}, x = - \frac{π + 2 πn - 2}{2}

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Frequently Asked Questions (FAQ)

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