What is the general solution for 1-sin(A)cos(A)=(sin(A)-cos(A))^2 ?

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

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1-sin(A)cos(A)=(sin(A)-cos(A))^2

1 - sin (A) cos (A) = (sin (A) - cos (A))^{2}

A = πn, A = \frac{π}{2} + πn

Degrees

A = 0^{\circ} + 18 0^{\circ} n, A = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

1 - sin (A) cos (A) = (sin (A) - cos (A))^{2}

Subtract

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

from both sides

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

Rewrite using trig identities

1 + cos (A) sin (A) - 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= 1 - 1 + \frac{sin ( 2 A )}{2}

1 - 1 + \frac{sin ( 2 A )}{2} = 0

1 - 1 = 0

\frac{sin ( 2 A )}{2} = 0

Multiply both sides by

2

\frac{sin ( 2 A )}{2} = 0

Multiply both sides by

2

\frac{2 sin ( 2 A )}{2} = 0 \cdot 2

Simplify

sin (2 A) = 0

sin (2 A) = 0

General solutions for

sin (2 A) = 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}

2 A = 0 + 2 πn, 2 A = π + 2 πn

2 A = 0 + 2 πn, 2 A = π + 2 πn

Solve

2 A = 0 + 2 πn : A = πn

2 A = 0 + 2 πn

0 + 2 πn = 2 πn

2 A = 2 πn

Divide both sides by

2

2 A = 2 πn

Divide both sides by

2

\frac{2 A}{2} = \frac{2 πn}{2}

Simplify

A = πn

A = πn

Solve

2 A = π + 2 πn : A = \frac{π}{2} + πn

2 A = π + 2 πn

Divide both sides by

2

2 A = π + 2 πn

Divide both sides by

2

\frac{2 A}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

A = \frac{π}{2} + πn

A = \frac{π}{2} + πn

A = πn, A = \frac{π}{2} + πn

sin (43. 5^{\circ}) = 1.5 \cdot sin (x)

cos (x) = \frac{- 1}{10}

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

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

- 1 + sin (x) = - 1

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