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5cos(x)+sqrt(2)=3cos(x),0<= x<= 2pi

Solución

5 cos (x) + 2 = 3 cos (x), 0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

Grados

x = 13 5^{\circ}, x = 22 5^{\circ}

Pasos de solución

5 cos (x) + 2 = 3 cos (x), 0 \leq x \leq 2 π

Usando el método de sustitución

5 cos (x) + 2 = 3 cos (x)

Sea:

cos (x) = u

5 u + 2 = 3 u

5 u + 2 = 3 u : u = - \frac{2}{2}

5 u + 2 = 3 u

Mover

2

al lado derecho

5 u + 2 = 3 u

Restar

2

de ambos lados

5 u + 2 - 2 = 3 u - 2

Simplificar

5 u = 3 u - 2

5 u = 3 u - 2

Mover

3 u

al lado izquierdo

5 u = 3 u - 2

Restar

3 u

de ambos lados

5 u - 3 u = 3 u - 2 - 3 u

Simplificar

2 u = - 2

2 u = - 2

Dividir ambos lados entre

2

2 u = - 2

Dividir ambos lados entre

2

\frac{2 u}{2} = \frac{- 2}{2}

Simplificar

u = - \frac{2}{2}

u = - \frac{2}{2}

Sustituir en la ecuación

u = cos (x)

cos (x) = - \frac{2}{2}

cos (x) = - \frac{2}{2}

cos (x) = - \frac{2}{2}, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{5 π}{4}

cos (x) = - \frac{2}{2}, 0 \leq x \leq 2 π

Soluciones generales para

cos (x) = - \frac{2}{2}

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

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}

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

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

Soluciones para el rango

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

Combinar toda las soluciones

x = \frac{3 π}{4}, x = \frac{5 π}{4}

2 tan (2 x + 10) = - 5.5

sin (t) = sin (2 t)

\frac{3}{2} = cos (x)

tan (x + \frac{π}{6}) = \frac{2}{3}

4 cos^{2} (x) + 4 sin (x) - 5 = 0