What is the general solution for 3+cos(4θ)=2-cos(4θ) ?

The general solution for 3+cos(4θ)=2-cos(4θ) is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2

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3+cos(4θ)=2-cos(4θ)

3 + cos (4 θ) = 2 - cos (4 θ)

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

Solution steps

3 + cos (4 θ) = 2 - cos (4 θ)

Solve by substitution

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \frac{1}{2} : θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

Combine all the solutions

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

5 sec (θ) = 4 sec (θ) + 2

4 sec (π x) + 6 = - 2

x, y - 1 = - 4 + 3 sin (4 x - \frac{π}{4})

64 = 81 + 25 - 90 (cos (C))

sin^{2} (a - \frac{π}{8}) = \frac{2}{3}

What is the general solution for 3+cos(4θ)=2-cos(4θ) ?
The general solution for 3+cos(4θ)=2-cos(4θ) is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2