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

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

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Popular >

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

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 113.54096 \dots^{\circ} n, x = 0^{\circ} - 113.54096 \dots^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

1 = tan (x^{2})

Switch sides

tan (x^{2}) = 1

tan (x^{2}) = 1

General solutions for

tan (x^{2}) = 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} = \frac{π}{4} + πn

x^{2} = \frac{π}{4} + πn

Solve

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

x^{2} = \frac{π}{4} + πn

For

x^{2} = f (a)

the solutions are

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

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

Simplify

\frac{π}{4} + πn : \frac{π + 4 πn}{2}

\frac{π}{4} + πn

Join

\frac{π}{4} + πn : \frac{π + 4 πn}{4}

\frac{π}{4} + πn

Convert element to fraction:

πn = \frac{πn 4}{4}

= \frac{π}{4} + \frac{πn \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}{4}

= \frac{π + πn \cdot 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{π + 4 πn}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{π + 4 πn}{2}

Simplify

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

- \frac{π}{4} + πn

Join

\frac{π}{4} + πn : \frac{π + 4 πn}{4}

\frac{π}{4} + πn

Convert element to fraction:

πn = \frac{πn 4}{4}

= \frac{π}{4} + \frac{πn \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}{4}

= - \frac{π + 4 πn}{4}

Simplify

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

\frac{π + πn \cdot 4}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{π + 4 πn}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{π + 4 πn}{2}

= - \frac{π + 4 πn}{2}

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

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

Popular Examples

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

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

5sin(2x)+3cos(2x)=0,0<= x<= 180

5 sin (2 x) + 3 cos (2 x) = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

cos(θ)=0.6798,0<= θ<= 360

cos (θ) = 0.6798, 0^{\circ} \leq θ \leq 36 0^{\circ}

(2sin(x)cos(x))/(cos(x))=1

\frac{2 sin ( x ) cos ( x )}{cos ( x )} = 1

csc(3θ)=sec(15)

csc (3 θ) = sec (1 5^{\circ})

Frequently Asked Questions (FAQ)

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