solvefor t,y^{11}+9y=18cos(3t)

Solution

solve for

t, y^{11} + 9 y = 18 cos (3 t)

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}, t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Solution steps

y^{11} + 9 y = 18 cos (3 t)

Switch sides

18 cos (3 t) = y^{11} + 9 y

Divide both sides by

18

cos (3 t) = \frac{y ^{11} + 9 y}{18}

Apply trig inverse properties

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn, 3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Solve

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn : t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Solve

3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn : t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}, t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

tan (x) = \frac{0.75}{6}

2 sec (x) - 2 = 0

(tan (x)) (sin^{2} (x) - a) (sec (x) - a) = 0

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sin (x) - 1 = 3 cos (x)