4cos(2θ)+16cos(θ)+4=9cos(θ)

Solución

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

Pasos de solución

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

Restar

9 cos (θ)

de ambos lados

4 cos (2 θ) + 7 cos (θ) + 4 = 0

Re-escribir usando identidades trigonométricas

7 cos (θ) + 8 cos^{2} (θ) = 0

Usando el método de sustitución

cos (θ) = 0, cos (θ) = - \frac{7}{8}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = - \frac{7}{8} : θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

Combinar toda las soluciones

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

Mostrar soluciones en forma decimal

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

sin (θ) = \frac{7 π}{3}

tan (x) \cdot cos^{2} (x) - tan (x) = 0

sin (x + π) - sin (x) + 2 = 0

3 tan (x) = - 3

2 (sin (x))^{2} - 5 cos (x) + 1 = 0