What is the general solution for 0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi ?

The general solution for 0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi

0 = cos^{2} (x) + sin^{2} (x) - 2 sin (x), 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Degrees

x = 3 0^{\circ}, x = 15 0^{\circ}

Solution steps

0 = cos^{2} (x) + sin^{2} (x) - 2 sin (x), 0 \leq x \leq 2 π

Switch sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - 2 sin (x) + 1

- 2 sin (x) + 1 = 0

Move

1

to the right side

- 2 sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

- 2 sin (x) = - 1

- 2 sin (x) = - 1

Divide both sides by

- 2

- 2 sin (x) = - 1

Divide both sides by

- 2

\frac{- 2 sin ( x )}{- 2} = \frac{- 1}{- 2}

Simplify

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (θ) = - \frac{7}{25}, sin (2 θ)

cos (θ) = \frac{9}{12}

csc (x) + cot (x) = - 1, 0 \leq x \leq 2 π

sin (x) + cos (x) = \frac{5}{7}, 9 0^{\circ} > x > 0^{\circ}

2 cos^{2} (θ) + cos (θ) - 1 = 0, \forall 0 \leq θ < 2 π

What is the general solution for 0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi ?
The general solution for 0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi is x= pi/6 ,x=(5pi)/6