5sin(2x)=2tan(30)

Solución

5 sin (2 x) = 2 tan (3 0^{\circ})

x = \frac{0.23304 \dots}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{0.23304 \dots}{2} + 18 0^{\circ} n

Pasos de solución

5 sin (2 x) = 2 tan (3 0^{\circ})

tan (3 0^{\circ}) = \frac{3}{3}

5 sin (2 x) = 2 \cdot \frac{3}{3}

Dividir ambos lados entre

5

sin (2 x) = \frac{2 3}{15}

Aplicar propiedades trigonométricas inversas

2 x = arcsin (\frac{2 3}{15}) + 36 0^{\circ} n, 2 x = 18 0^{\circ} - arcsin (\frac{2 3}{15}) + 36 0^{\circ} n

Resolver

2 x = arcsin (\frac{2 3}{15}) + 36 0^{\circ} n : x = \frac{arcsin ( \frac{2 3}{15} )}{2} + 18 0^{\circ} n

Resolver

2 x = 18 0^{\circ} - arcsin (\frac{2 3}{15}) + 36 0^{\circ} n : x = 9 0^{\circ} - \frac{arcsin ( \frac{2 3}{15} )}{2} + 18 0^{\circ} n

x = \frac{arcsin ( \frac{2 3}{15} )}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{arcsin ( \frac{2 3}{15} )}{2} + 18 0^{\circ} n

Mostrar soluciones en forma decimal

x = \frac{0.23304 \dots}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{0.23304 \dots}{2} + 18 0^{\circ} n

6 sin^{2} (x) + cos (x) - 4 = 0

sin (2 x) sin (x) + cos (x) = 0

sin (x) - 2 cos (x) = 0

sec (x) = csc (x)

cos (x) = cos (x) sin (x)