What is the general solution for 8cos(2x-5)=3 ?

The general solution for 8cos(2x-5)=3 is x=pin+5/2+(1.18639…)/2 ,x=pi+pin+5/2-(1.18639…)/2

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8cos(2x-5)=3

8 cos (2 x - 5) = 3

x = πn + \frac{5}{2} + \frac{1.18639 \dots}{2}, x = π + πn + \frac{5}{2} - \frac{1.18639 \dots}{2}

Solution steps

8 cos (2 x - 5) = 3

Divide both sides by

8

cos (2 x - 5) = \frac{3}{8}

Apply trig inverse properties

2 x - 5 = arccos (\frac{3}{8}) + 2 πn, 2 x - 5 = 2 π - arccos (\frac{3}{8}) + 2 πn

Solve

2 x - 5 = arccos (\frac{3}{8}) + 2 πn : x = πn + \frac{5}{2} + \frac{arccos ( \frac{3}{8} )}{2}

Solve

2 x - 5 = 2 π - arccos (\frac{3}{8}) + 2 πn : x = π + πn + \frac{5}{2} - \frac{arccos ( \frac{3}{8} )}{2}

x = πn + \frac{5}{2} + \frac{arccos ( \frac{3}{8} )}{2}, x = π + πn + \frac{5}{2} - \frac{arccos ( \frac{3}{8} )}{2}

Show solutions in decimal form

x = πn + \frac{5}{2} + \frac{1.18639 \dots}{2}, x = π + πn + \frac{5}{2} - \frac{1.18639 \dots}{2}

\frac{116}{sin ( 9 0 ^{\circ} )} = \frac{4}{sin ( x )}

\frac{3}{cos ^{2} ( x )} = \frac{7 + 4}{cot ( x )}

sin (x) = 0.01

sin (x) - sin (2 x) = sin^{3} (x)

- sin^{2} (x) = cos (2 x)

What is the general solution for 8cos(2x-5)=3 ?
The general solution for 8cos(2x-5)=3 is x=pin+5/2+(1.18639…)/2 ,x=pi+pin+5/2-(1.18639…)/2