1/(csc^2(a))=1-sin^2(a)

Solución

\frac{1}{csc ^{2} ( a )} = 1 - sin^{2} (a)

a = \frac{π}{4} + 2 πn, a = \frac{3 π}{4} + 2 πn, a = \frac{5 π}{4} + 2 πn, a = \frac{7 π}{4} + 2 πn

Pasos de solución

\frac{1}{csc ^{2} ( a )} = 1 - sin^{2} (a)

Restar

1 - sin^{2} (a)

de ambos lados

\frac{1}{csc ^{2} ( a )} - 1 + sin^{2} (a) = 0

Simplificar

\frac{1}{csc ^{2} ( a )} - 1 + sin^{2} (a) : \frac{1 - csc ^{2} ( a ) + sin ^{2} ( a ) csc ^{2} ( a )}{csc ^{2} ( a )}

\frac{1 - csc ^{2} ( a ) + sin ^{2} ( a ) csc ^{2} ( a )}{csc ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - csc^{2} (a) + sin^{2} (a) csc^{2} (a) = 0

Re-escribir usando identidades trigonométricas

2 - csc^{2} (a) = 0

Usando el método de sustitución

csc (a) = 2, csc (a) = - 2

csc (a) = 2 : a = \frac{π}{4} + 2 πn, a = \frac{3 π}{4} + 2 πn

csc (a) = - 2 : a = \frac{5 π}{4} + 2 πn, a = \frac{7 π}{4} + 2 πn

Combinar toda las soluciones

a = \frac{π}{4} + 2 πn, a = \frac{3 π}{4} + 2 πn, a = \frac{5 π}{4} + 2 πn, a = \frac{7 π}{4} + 2 πn

5 cos (3 x) = - 3

2 sin^{3} (x) + sin^{2} (x) = 0

2 sin^{2} (θ) = sin (θ), 0 \leq θ \leq 2 π

cos (2 x) = cos (4 x)

14 sin^{2} (x) - 7 sin (x) = 0