What is the general solution for sin(x^2+1)-cos(x^2+1)=0 ?

The general solution for sin(x^2+1)-cos(x^2+1)=0 is x=(sqrt(pi+4pin-4))/2 ,x=-(sqrt(pi+4pin-4))/2

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

Solution

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

Solution

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

Degrees

x = 0^{\circ} + 98.02420 \dots^{\circ} n, x = 0^{\circ} - 98.02420 \dots^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

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

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

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

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

General solutions for

tan (x^{2} + 1) = 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^{2} + 1 = \frac{π}{4} + πn

x^{2} + 1 = \frac{π}{4} + πn

Solve

x^{2} + 1 = \frac{π}{4} + πn : x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

x^{2} + 1 = \frac{π}{4} + πn

Move

1

to the right side

x^{2} + 1 = \frac{π}{4} + πn

Subtract

1

from both sides

x^{2} + 1 - 1 = \frac{π}{4} + πn - 1

Simplify

x^{2} = \frac{π}{4} + πn - 1

x^{2} = \frac{π}{4} + πn - 1

For

x^{2} = f (a)

the solutions are

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

x = \frac{π}{4} + πn - 1, x = - \frac{π}{4} + πn - 1

Simplify

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{2}

\frac{π}{4} + πn - 1

Join

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{4}

\frac{π}{4} + πn - 1

Convert element to fraction:

πn = \frac{πn 4}{4}, 1 = \frac{1 \cdot 4}{4}

= \frac{π}{4} + \frac{πn \cdot 4}{4} - \frac{1 \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π + πn \cdot 4 - 1 \cdot 4}{4}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{π + 4 πn - 4}{4}

= \frac{π + πn \cdot 4 - 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{π + 4 πn - 4}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{π + 4 πn - 4}{2}

Simplify

- \frac{π}{4} + πn - 1 : - \frac{π + 4 πn - 4}{2}

- \frac{π}{4} + πn - 1

Join

\frac{π}{4} + πn - 1 : \frac{π + 4 πn - 4}{4}

\frac{π}{4} + πn - 1

Convert element to fraction:

πn = \frac{πn 4}{4}, 1 = \frac{1 \cdot 4}{4}

= \frac{π}{4} + \frac{πn \cdot 4}{4} - \frac{1 \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π + πn \cdot 4 - 1 \cdot 4}{4}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{π + 4 πn - 4}{4}

= - \frac{π + 4 πn - 4}{4}

Simplify

\frac{π + πn \cdot 4 - 4}{4} : \frac{π + 4 πn - 4}{2}

\frac{π + πn \cdot 4 - 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{π + 4 πn - 4}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{π + 4 πn - 4}{2}

= - \frac{π + 4 πn - 4}{2}

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

x = \frac{π + 4 πn - 4}{2}, x = - \frac{π + 4 πn - 4}{2}

Popular Examples

cos(θ)=(sqrt(2))/3

cos (θ) = \frac{2}{3}

sin(8x)=0

sin (8 x) = 0

sin(θ/2)=(sqrt(2))/2

sin (\frac{θ}{2}) = \frac{2}{2}

9tan(3x)=9

9 tan (3 x) = 9

sin(θ)= 40/41

sin (θ) = \frac{40}{41}

Frequently Asked Questions (FAQ)

What is the general solution for sin(x^2+1)-cos(x^2+1)=0 ?
The general solution for sin(x^2+1)-cos(x^2+1)=0 is x=(sqrt(pi+4pin-4))/2 ,x=-(sqrt(pi+4pin-4))/2