What is the general solution for 7/4 cos(3θ)-1=0 ?

The general solution for 7/4 cos(3θ)-1=0 is θ=(0.96255…)/3+(2pin)/3 ,θ=(2pi)/3-(0.96255…)/3+(2pin)/3

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7/4 cos(3θ)-1=0

\frac{7}{4} cos (3 θ) - 1 = 0

θ = \frac{0.96255 \dots}{3} + \frac{2 πn}{3}, θ = \frac{2 π}{3} - \frac{0.96255 \dots}{3} + \frac{2 πn}{3}

Solution steps

\frac{7}{4} cos (3 θ) - 1 = 0

Move

1

to the right side

\frac{7}{4} cos (3 θ) = 1

Multiply both sides by

4

7 cos (3 θ) = 4

Divide both sides by

7

cos (3 θ) = \frac{4}{7}

Apply trig inverse properties

3 θ = arccos (\frac{4}{7}) + 2 πn, 3 θ = 2 π - arccos (\frac{4}{7}) + 2 πn

Solve

3 θ = arccos (\frac{4}{7}) + 2 πn : θ = \frac{arccos ( \frac{4}{7} )}{3} + \frac{2 πn}{3}

Solve

3 θ = 2 π - arccos (\frac{4}{7}) + 2 πn : θ = \frac{2 π}{3} - \frac{arccos ( \frac{4}{7} )}{3} + \frac{2 πn}{3}

θ = \frac{arccos ( \frac{4}{7} )}{3} + \frac{2 πn}{3}, θ = \frac{2 π}{3} - \frac{arccos ( \frac{4}{7} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

θ = \frac{0.96255 \dots}{3} + \frac{2 πn}{3}, θ = \frac{2 π}{3} - \frac{0.96255 \dots}{3} + \frac{2 πn}{3}

cos (x) = sin (\frac{7 π}{6}), 0 \leq x \leq 2 π

cos (2 x) = \frac{( 3 k - 3 )}{8}

1.5 = 2.75 cos (30 (x - 4)) + 3.25

sin (x) = (\frac{5}{12})

sin (x) = \frac{97}{270}

What is the general solution for 7/4 cos(3θ)-1=0 ?
The general solution for 7/4 cos(3θ)-1=0 is θ=(0.96255…)/3+(2pin)/3 ,θ=(2pi)/3-(0.96255…)/3+(2pin)/3