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

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

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

Solution steps

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

Rewrite using trig identities

(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

The solution to the quadratic equation is:

sin (x + \frac{π}{4}) = 0

General solutions for

sin (x + \frac{π}{4}) = 0

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}

Solve

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

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

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

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

7.5 = 5 sin (\frac{π}{2} (x - 2)) + 5

u, y = ccos h \cdot sin (u)

2 cos (θ) + 23 = 3

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