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

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

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

Solution

cos (x + \frac{π}{2}) = sin (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (x + \frac{π}{2}) = sin (x)

Rewrite using trig identities

cos (x + \frac{π}{2}) = sin (x)

Rewrite using trig identities

cos (x + \frac{π}{2})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

Simplify

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

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

cos (x) cos (\frac{π}{2}) = 0

cos (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (x) sin (\frac{π}{2}) = sin (x)

sin (x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1 \cdot sin (x)

Multiply:

sin (x) \cdot 1 = sin (x)

= sin (x)

= 0 - sin (x)

0 - sin (x) = - sin (x)

= - sin (x)

= - sin (x)

- sin (x) = sin (x)

- sin (x) = sin (x)

Subtract

sin (x)

from both sides

- 2 sin (x) = 0

Divide both sides by

- 2

- 2 sin (x) = 0

Divide both sides by

- 2

\frac{- 2 sin ( x )}{- 2} = \frac{0}{- 2}

Simplify

sin (x) = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Popular Examples

cos(x)tan(x)+cos(x)=0

cos (x) tan (x) + cos (x) = 0

2sin^2(2θ)=1+sin(2θ)

2 sin^{2} (2 θ) = 1 + sin (2 θ)

sin(2x)-cos(x)=0,0<= x<= 2pi

sin (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

csc(θ)=-2/(sqrt(3))

csc (θ) = - \frac{2}{3}

2cos(x/2)+sqrt(3)=0

2 cos (\frac{x}{2}) + 3 = 0

Frequently Asked Questions (FAQ)

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