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

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

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

Solution

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

Solution

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

Degrees

x = - 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

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

sin (x) + cos (x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

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

Use the Angle Sum identity:

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

= 2 sin (x + \frac{π}{4})

= (2 sin (x + \frac{π}{4}))^{2}

(2 sin (x + \frac{π}{4}))^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

2 sin (x + \frac{π}{4}) = 0 : sin (x + \frac{π}{4}) = 0

2 sin (x + \frac{π}{4}) = 0

Divide both sides by

2

2 sin (x + \frac{π}{4}) = 0

Divide both sides by

2

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

Simplify

sin (x + \frac{π}{4}) = 0

sin (x + \frac{π}{4}) = 0

The solution to the quadratic equation is:

sin (x + \frac{π}{4}) = 0

General solutions for

sin (x + \frac{π}{4}) = 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 + \frac{π}{4} = 0 + 2 πn, x + \frac{π}{4} = π + 2 πn

x + \frac{π}{4} = 0 + 2 πn, x + \frac{π}{4} = π + 2 πn

Solve

x + \frac{π}{4} = 0 + 2 πn : x = 2 πn - \frac{π}{4}

x + \frac{π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

x + \frac{π}{4} = 2 πn

Move

\frac{π}{4}

to the right side

x + \frac{π}{4} = 2 πn

Subtract

\frac{π}{4}

from both sides

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

Simplify

x = 2 πn - \frac{π}{4}

x = 2 πn - \frac{π}{4}

Solve

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

x + \frac{π}{4} = π + 2 πn

Move

\frac{π}{4}

to the right side

x + \frac{π}{4} = π + 2 πn

Subtract

\frac{π}{4}

from both sides

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

Simplify

x = π + 2 πn - \frac{π}{4}

x = π + 2 πn - \frac{π}{4}

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

Popular Examples

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tan (θ) = \frac{3}{3}, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

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7.5 = 5 sin (\frac{π}{2} (x - 2)) + 5

solvefor u,y=ccosh*sin(u)solve for

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2 cos (θ) + 23 = 3

Frequently Asked Questions (FAQ)

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