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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

1 - 1 = 0

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

Multiply both sides by

2

sin (2 A) = 0

General solutions for

sin (2 A) = 0

2 A = 0 + 2 πn, 2 A = π + 2 πn

Solve

2 A = 0 + 2 πn : A = πn

Solve

2 A = 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