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

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

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sin^2(A)+cos^2(A)+sin(A)-2=0

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

A = \frac{π}{2} + 2 πn

Degrees

A = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

sin^{2} (A) + cos^{2} (A) + sin (A) - 2

Use the Pythagorean identity:

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

= - 2 + sin (A) + 1

- 2 + sin (A) + 1 = 0

Group like terms

sin (A) - 2 + 1 = 0

Add/Subtract the numbers:

- 2 + 1 = - 1

sin (A) - 1 = 0

Move

1

to the right side

sin (A) - 1 = 0

Add

1

to both sides

sin (A) - 1 + 1 = 0 + 1

Simplify

sin (A) = 1

sin (A) = 1

General solutions for

sin (A) = 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}

A = \frac{π}{2} + 2 πn

A = \frac{π}{2} + 2 πn

2 - 2 cos (x) = - 6 cos (x)

2 sin^{2} (u) = 1 + sin (u), 0 \leq t < 2 π

tan (3 y + 60) tan (2 y + 5) = 1

1 8^{2} = 8^{2} + 2 0^{2} - 2 \cdot 8 \cdot 20 \cdot cos (x)

cot (θ) = - \frac{π}{6}

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