What is the general solution for cos(2+3*x)-cos(-0.5)=0 ?

The general solution for cos(2+3*x)-cos(-0.5)=0 is x=(2pin)/3-1/2 ,x=-5/6+(2pi)/3+(2pin)/3

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cos(2+3*x)-cos(-0.5)=0

cos (2 + 3 \cdot x) - cos (- 0.5) = 0

x = \frac{2 πn}{3} - \frac{1}{2}, x = - \frac{5}{6} + \frac{2 π}{3} + \frac{2 πn}{3}

Solution steps

cos (2 + 3 x) - cos (- 0.5) = 0

cos (- 0.5) = cos (\frac{1}{2})

cos (2 + 3 x) - cos (\frac{1}{2}) = 0

Move

cos (\frac{1}{2})

to the right side

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

Apply trig inverse properties

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

Solve

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

Solve

2 + 3 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn : x = - \frac{5}{6} + \frac{2 π}{3} + \frac{2 πn}{3}

x = \frac{2 πn}{3} - \frac{1}{2}, x = - \frac{5}{6} + \frac{2 π}{3} + \frac{2 πn}{3}

2 cos (θ) = 1, 0 \leq θ \leq 2 π

arctan (x + 1) + arctan (x - 1) = arctan (\frac{8}{31})

4 cos (2 θ) + 19 = - 22 cos (θ) + 6

8 cos^{2} (x) + 16 cos (x) + 8 = 0

cos (x) = - 0.35

What is the general solution for cos(2+3*x)-cos(-0.5)=0 ?
The general solution for cos(2+3*x)-cos(-0.5)=0 is x=(2pin)/3-1/2 ,x=-5/6+(2pi)/3+(2pin)/3