What is the general solution for sin(x)+pi/2 =cos(x) ?

The general solution for sin(x)+pi/2 =cos(x) is No Solution for x\in\mathbb{R}

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sin(x)+pi/2 =cos(x)

Solution

sin (x) + \frac{π}{2} = cos (x)

Solution

No Solution for x \in R

Solution steps

sin (x) + \frac{π}{2} = cos (x)

Square both sides

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

Subtract

cos^{2} (x)

from both sides

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x) = 0

Simplify

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x) : \frac{4 sin ^{2} ( x ) + 4 π sin ( x ) + π ^{2} - 4 cos ^{2} ( x )}{4}

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x)

Convert element to fraction:

sin^{2} (x) = \frac{sin ^{2} ( x ) 4}{4}, π sin (x) = \frac{π sin ( x ) 4}{4}, cos^{2} (x) = \frac{cos ^{2} ( x ) 4}{4}

= \frac{sin ^{2} ( x ) \cdot 4}{4} + \frac{π sin ( x ) \cdot 4}{4} + \frac{π ^{2}}{4} - \frac{cos ^{2} ( x ) \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( x ) \cdot 4 + π sin ( x ) \cdot 4 + π ^{2} - cos ^{2} ( x ) \cdot 4}{4}

\frac{4 sin ^{2} ( x ) + 4 π sin ( x ) + π ^{2} - 4 cos ^{2} ( x )}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (x) + 4 π sin (x) + π^{2} - 4 cos^{2} (x) = 0

Rewrite using trig identities

π^{2} - 4 cos^{2} (x) + 4 sin^{2} (x) + 4 sin (x) π

Use the Pythagorean identity:

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

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

= π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π

Simplify

π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π : π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π

= π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 π sin (x)

Expand

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= π^{2} - 4 + 4 sin^{2} (x) + 4 sin^{2} (x) + 4 sin (x) π

Add similar elements:

4 sin^{2} (x) + 4 sin^{2} (x) = 8 sin^{2} (x)

= π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

= π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

- 4 + π^{2} + 8 sin^{2} (x) + 4 sin (x) π = 0

Solve by substitution

- 4 + π^{2} + 8 sin^{2} (x) + 4 sin (x) π = 0

Let:

sin (x) = u

- 4 + π^{2} + 8 u^{2} + 4 u π = 0

- 4 + π^{2} + 8 u^{2} + 4 u π = 0 : u = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, u = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

- 4 + π^{2} + 8 u^{2} + 4 u π = 0

Write in the standard form

a x^{2} + b x + c = 0

8 u^{2} + 4 π u - 4 + π^{2} = 0

Solve with the quadratic formula

8 u^{2} + 4 π u - 4 + π^{2} = 0

Quadratic Equation Formula:

For

a = 8, b = 4 π, c = - 4 + π^{2}

u_{1, 2} = \frac{- 4 π \pm ( 4 π ) ^{2} - 4 \cdot 8 ( - 4 + π ^{2} )}{2 \cdot 8}

u_{1, 2} = \frac{- 4 π \pm ( 4 π ) ^{2} - 4 \cdot 8 ( - 4 + π ^{2} )}{2 \cdot 8}

Simplify

(4 π)^{2} - 4 \cdot 8 (- 4 + π^{2}) : i 16 π^{2} - 128

(4 π)^{2} - 4 \cdot 8 (- 4 + π^{2})

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} π^{2} - 4 \cdot 8 (π^{2} - 4)

Multiply the numbers:

4 \cdot 8 = 32

= 4^{2} π^{2} - 32 (π^{2} - 4)

Apply imaginary number rule:

- a = i a

= i 32 (π^{2} - 4) - 4^{2} π^{2}

- 4^{2} π^{2} + 32 (- 4 + π^{2}) = 16 π^{2} - 128

- 4^{2} π^{2} + 32 (- 4 + π^{2})

4^{2} = 16

= - 16 π^{2} + 32 (π^{2} - 4)

Expand

- 16 π^{2} + 32 (- 4 + π^{2}) : 16 π^{2} - 128

- 16 π^{2} + 32 (- 4 + π^{2})

Expand

32 (- 4 + π^{2}) : - 128 + 32 π^{2}

32 (- 4 + π^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 32, b = - 4, c = π^{2}

= 32 (- 4) + 32 π^{2}

Apply minus-plus rules

+ (- a) = - a

= - 32 \cdot 4 + 32 π^{2}

Multiply the numbers:

32 \cdot 4 = 128

= - 128 + 32 π^{2}

= - 16 π^{2} - 128 + 32 π^{2}

Simplify

- 16 π^{2} - 128 + 32 π^{2} : 16 π^{2} - 128

- 16 π^{2} - 128 + 32 π^{2}

Group like terms

= - 16 π^{2} + 32 π^{2} - 128

Add similar elements:

- 16 π^{2} + 32 π^{2} = 16 π^{2}

= 16 π^{2} - 128

= 16 π^{2} - 128

= 16 π^{2} - 128

= i 16 π^{2} - 128

u_{1, 2} = \frac{- 4 π \pm i 16 π ^{2} - 128}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8}, u_{2} = \frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8}

u = \frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8} : - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}

\frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4 π + i 16 π ^{2} - 128}{16}

Rewrite

\frac{- 4 π + i 16 π ^{2} - 128}{16}

in standard complex form:

- \frac{π}{4} + \frac{16 π ^{2} - 128}{16} i

\frac{- 4 π + i 16 π ^{2} - 128}{16}

Apply the fraction rule:

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

\frac{- 4 π + i 16 π ^{2} - 128}{16} = - \frac{4 π}{16} + \frac{i 16 π ^{2} - 128}{16}

= - \frac{4 π}{16} + \frac{i 16 π ^{2} - 128}{16}

Cancel

\frac{4 π}{16} : \frac{π}{4}

\frac{4 π}{16}

Cancel the common factor:

4

= \frac{π}{4}

= - \frac{π}{4} + \frac{i 16 π ^{2} - 128}{16}

= - \frac{π}{4} + \frac{16 π ^{2} - 128}{16} i

u = \frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8} : - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

\frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4 π - i 16 π ^{2} - 128}{16}

Rewrite

\frac{- 4 π - i 16 π ^{2} - 128}{16}

in standard complex form:

- \frac{π}{4} - \frac{16 π ^{2} - 128}{16} i

\frac{- 4 π - i 16 π ^{2} - 128}{16}

Apply the fraction rule:

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

\frac{- 4 π - i 16 π ^{2} - 128}{16} = - \frac{4 π}{16} - \frac{i 16 π ^{2} - 128}{16}

= - \frac{4 π}{16} - \frac{i 16 π ^{2} - 128}{16}

Cancel

\frac{4 π}{16} : \frac{π}{4}

\frac{4 π}{16}

Cancel the common factor:

4

= \frac{π}{4}

= - \frac{π}{4} - \frac{i 16 π ^{2} - 128}{16}

= - \frac{π}{4} - \frac{16 π ^{2} - 128}{16} i

The solutions to the quadratic equation are:

u = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, u = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

Substitute back

u = sin (x)

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16} :

No Solution

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}

No Solution

sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16} :

No Solution

sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (x) + \frac{π}{2} = cos (x)

Remove the ones that don't agree with the equation.

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+pi/2 =cos(x) ?
The general solution for sin(x)+pi/2 =cos(x) is No Solution for x\in\mathbb{R}