What is the general solution for 3cos(2x+1)=2 ?

The general solution for 3cos(2x+1)=2 is x=pin-1/2+(0.84106…)/2 ,x=pi+pin-1/2-(0.84106…)/2

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3cos(2x+1)=2

3 cos (2 x + 1) = 2

x = πn - \frac{1}{2} + \frac{0.84106 \dots}{2}, x = π + πn - \frac{1}{2} - \frac{0.84106 \dots}{2}

Solution steps

3 cos (2 x + 1) = 2

Divide both sides by

3

cos (2 x + 1) = \frac{2}{3}

Apply trig inverse properties

2 x + 1 = arccos (\frac{2}{3}) + 2 πn, 2 x + 1 = 2 π - arccos (\frac{2}{3}) + 2 πn

Solve

2 x + 1 = arccos (\frac{2}{3}) + 2 πn : x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

Solve

2 x + 1 = 2 π - arccos (\frac{2}{3}) + 2 πn : x = π + πn - \frac{1}{2} - \frac{arccos ( \frac{2}{3} )}{2}

x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}, x = π + πn - \frac{1}{2} - \frac{arccos ( \frac{2}{3} )}{2}

Show solutions in decimal form

x = πn - \frac{1}{2} + \frac{0.84106 \dots}{2}, x = π + πn - \frac{1}{2} - \frac{0.84106 \dots}{2}

sin (t) = \frac{3}{7}, cos (t)

tan (x) = \frac{16}{12}

sin (4 x + 10) = sin (9 0^{\circ})

y, - \frac{1}{2} cot (y) = \frac{t ^{2}}{+ 2} + C

3 sec (2 θ) + 2 = 0

What is the general solution for 3cos(2x+1)=2 ?
The general solution for 3cos(2x+1)=2 is x=pin-1/2+(0.84106…)/2 ,x=pi+pin-1/2-(0.84106…)/2