What is the general solution for (sin(x)+1)^2+cos^2(x)=2 ?

The general solution for (sin(x)+1)^2+cos^2(x)=2 is x=2pin,x=pi+2pin

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

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

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

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

Subtract

2

from both sides

sin^{2} (x) + cos^{2} (x) + 2 sin (x) - 1 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - 1 + 2 sin (x) + 1

- 1 + 2 sin (x) + 1 = 0

Group like terms

2 sin (x) - 1 + 1 = 0

- 1 + 1 = 0

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

12 sin^{2} (x) - cos (x) - 6 = 0

tan (\frac{x}{2}) - 1 = 0

sin (θ) = - cos (θ)

cot^{2} (θ) + csc (θ) = 1

5 cos^{2} (θ) - 2 cos (θ) = 0

What is the general solution for (sin(x)+1)^2+cos^2(x)=2 ?
The general solution for (sin(x)+1)^2+cos^2(x)=2 is x=2pin,x=pi+2pin