What is the general solution for 2sin(4x)-1=0,0<x<= 2 ?

The general solution for 2sin(4x)-1=0,0<x<= 2 is x= pi/(24),x=(5pi)/(24),x=(13pi)/(24)

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2sin(4x)-1=0,0<x<= 2

2 sin (4 x) - 1 = 0, 0 < x \leq 2

x = \frac{π}{24}, x = \frac{5 π}{24}, x = \frac{13 π}{24}

Solution steps

2 sin (4 x) - 1 = 0, 0 < x \leq 2

Move

1

to the right side

2 sin (4 x) = 1

Divide both sides by

2

sin (4 x) = \frac{1}{2}

General solutions for

sin (4 x) = \frac{1}{2}

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

Solve

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

Solve

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

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

Solutions for the range

0 < x \leq 2

x = \frac{π}{24}, x = \frac{5 π}{24}, x = \frac{13 π}{24}

5 cos (θ) = - 5 sin (- θ)

cos (x) - 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

sec^{2} (x) + 6 tan (x) = 8

tan (θ) = - \frac{3}{8}

i, arcsin (sin (x)) = x - 2 π

What is the general solution for 2sin(4x)-1=0,0<x<= 2 ?
The general solution for 2sin(4x)-1=0,0<x<= 2 is x= pi/(24),x=(5pi)/(24),x=(13pi)/(24)