What is the general solution for 2cos(4x)+sqrt(3)=0 ?

The general solution for 2cos(4x)+sqrt(3)=0 is x=(5pi}{24}+(pin)/2 ,x=(7pi)/(24)+\frac{pin)/2

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2cos(4x)+sqrt(3)=0

2 cos (4 x) + 3 = 0

x = \frac{5 π}{24} + \frac{πn}{2}, x = \frac{7 π}{24} + \frac{πn}{2}

Solution steps

2 cos (4 x) + 3 = 0

Move

3

to the right side

2 cos (4 x) = - 3

Divide both sides by

2

cos (4 x) = - \frac{3}{2}

General solutions for

cos (4 x) = - \frac{3}{2}

4 x = \frac{5 π}{6} + 2 πn, 4 x = \frac{7 π}{6} + 2 πn

Solve

4 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{24} + \frac{πn}{2}

Solve

4 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{24} + \frac{πn}{2}

x = \frac{5 π}{24} + \frac{πn}{2}, x = \frac{7 π}{24} + \frac{πn}{2}

cos (θ) = 0.4

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

2 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

sin (θ) = 0.82

tan (x) = \frac{1}{10}

What is the general solution for 2cos(4x)+sqrt(3)=0 ?
The general solution for 2cos(4x)+sqrt(3)=0 is x=(5pi}{24}+(pin)/2 ,x=(7pi)/(24)+\frac{pin)/2