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

The general solution for 1=cos^2(α)-2cos(α)sin(a)+sin^2(α) is α= pi/2+2pin,α=(3pi)/2+2pin

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1=cos^2(α)-2cos(α)sin(a)+sin^2(α)

1 = cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

α = \frac{π}{2} + 2 πn, α = \frac{3 π}{2} + 2 πn

Degrees

α = 9 0^{\circ} + 36 0^{\circ} n, α = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

1 = cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

Switch sides

cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α) = 1

Rewrite using trig identities

cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

Use the Pythagorean identity:

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

= - 2 cos (α) sin (a) + 1

- 2 cos (α) sin (a) + 1 = 1

Subtract

1

from both sides

- 2 cos (α) sin (a) + 1 - 1 = 1 - 1

Simplify

- 2 cos (α) sin (a) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

cos (α) = 0

General solutions for

cos (α) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

α = \frac{π}{2} + 2 πn, α = \frac{3 π}{2} + 2 πn

α = \frac{π}{2} + 2 πn, α = \frac{3 π}{2} + 2 πn

4 sin (θ) = 1

sin (x + \frac{π}{2}) = 0.6

4 cot (x) + 1 = 1 + 2 cot (x)

\frac{2}{3} = sin (θ) - 0.28 cos (θ)

2 sin (x) = 3 - cos (x), 0 \leq x \leq 2 π

What is the general solution for 1=cos^2(α)-2cos(α)sin(a)+sin^2(α) ?
The general solution for 1=cos^2(α)-2cos(α)sin(a)+sin^2(α) is α= pi/2+2pin,α=(3pi)/2+2pin