What is the general solution for 6=10cos(4*o) ?

The general solution for 6=10cos(4*o) is o=(0.92729…)/4+(pin)/2 ,o= pi/2-(0.92729…)/4+(pin)/2

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6=10cos(4*o)

6 = 10 cos (4 \cdot o)

o = \frac{0.92729 \dots}{4} + \frac{πn}{2}, o = \frac{π}{2} - \frac{0.92729 \dots}{4} + \frac{πn}{2}

Solution steps

6 = 10 cos (4 o)

Switch sides

10 cos (4 o) = 6

Divide both sides by

10

cos (4 o) = \frac{3}{5}

Apply trig inverse properties

4 o = arccos (\frac{3}{5}) + 2 πn, 4 o = 2 π - arccos (\frac{3}{5}) + 2 πn

Solve

4 o = arccos (\frac{3}{5}) + 2 πn : o = \frac{arccos ( \frac{3}{5} )}{4} + \frac{πn}{2}

Solve

4 o = 2 π - arccos (\frac{3}{5}) + 2 πn : o = \frac{π}{2} - \frac{arccos ( \frac{3}{5} )}{4} + \frac{πn}{2}

o = \frac{arccos ( \frac{3}{5} )}{4} + \frac{πn}{2}, o = \frac{π}{2} - \frac{arccos ( \frac{3}{5} )}{4} + \frac{πn}{2}

Show solutions in decimal form

o = \frac{0.92729 \dots}{4} + \frac{πn}{2}, o = \frac{π}{2} - \frac{0.92729 \dots}{4} + \frac{πn}{2}

2 cos (2 θ) + 5 cos (θ) = 4

y, x = sin (\frac{1}{2} y)

sin^{2} (θ) - sin (θ) = 2

x, sin (x) cos (x) = 0

\frac{2}{3} = \frac{sin ( 13 5 ^{\circ} )}{sin ( x )}

What is the general solution for 6=10cos(4*o) ?
The general solution for 6=10cos(4*o) is o=(0.92729…)/4+(pin)/2 ,o= pi/2-(0.92729…)/4+(pin)/2