What is the general solution for cos(4t)=-3/4 ,0<= t<= 2pi ?

The general solution for cos(4t)=-3/4 ,0<= t<= 2pi is

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cos(4t)=-3/4 ,0<= t<= 2pi

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

t = \frac{2.41885 \dots}{4}, t = \frac{2.41885 \dots + 2 π}{4}, t = \frac{- 2.41885 \dots + 2 π}{4}, t = \frac{2.41885 \dots + 4 π}{4}, t = \frac{- 2.41885 \dots + 4 π}{4}, t = \frac{2.41885 \dots + 6 π}{4}, t = \frac{- 2.41885 \dots + 6 π}{4}, t = \frac{- 2.41885 \dots + 8 π}{4}

Solution steps

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

Apply trig inverse properties

4 t = arccos (- \frac{3}{4}) + 2 πn, 4 t = - arccos (- \frac{3}{4}) + 2 πn

Solve

4 t = arccos (- \frac{3}{4}) + 2 πn : t = \frac{arccos ( - \frac{3}{4} )}{4} + \frac{πn}{2}

Solve

4 t = - arccos (- \frac{3}{4}) + 2 πn : t = - \frac{arccos ( - \frac{3}{4} )}{4} + \frac{πn}{2}

t = \frac{arccos ( - \frac{3}{4} )}{4} + \frac{πn}{2}, t = - \frac{arccos ( - \frac{3}{4} )}{4} + \frac{πn}{2}

Solutions for the range

0 \leq t \leq 2 π

t = \frac{arccos ( - \frac{3}{4} )}{4}, t = \frac{arccos ( - \frac{3}{4} ) + 2 π}{4}, t = \frac{- arccos ( - \frac{3}{4} ) + 2 π}{4}, t = \frac{arccos ( - \frac{3}{4} ) + 4 π}{4}, t = \frac{- arccos ( - \frac{3}{4} ) + 4 π}{4}, t = \frac{arccos ( - \frac{3}{4} ) + 6 π}{4}, t = \frac{- arccos ( - \frac{3}{4} ) + 6 π}{4}, t = \frac{- arccos ( - \frac{3}{4} ) + 8 π}{4}

Show solutions in decimal form

t = \frac{2.41885 \dots}{4}, t = \frac{2.41885 \dots + 2 π}{4}, t = \frac{- 2.41885 \dots + 2 π}{4}, t = \frac{2.41885 \dots + 4 π}{4}, t = \frac{- 2.41885 \dots + 4 π}{4}, t = \frac{2.41885 \dots + 6 π}{4}, t = \frac{- 2.41885 \dots + 6 π}{4}, t = \frac{- 2.41885 \dots + 8 π}{4}

cos (x) = 0.71

cos (x) = 0.28

2 sin (θ) - 3 = - 2

cos (x) + 2 sec (x) = - 3

2 sin^{2} (x) = 1, 0 \leq x \leq 2 π

What is the general solution for cos(4t)=-3/4 ,0<= t<= 2pi ?
The general solution for cos(4t)=-3/4 ,0<= t<= 2pi is