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

The general solution for 4sin(x)cos(x)-1=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin

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4sin(x)cos(x)-1=0

4 sin (x) cos (x) - 1 = 0

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solution steps

4 sin (x) cos (x) - 1 = 0

Rewrite using trig identities

- 1 + 4 \cdot \frac{sin ( 2 x )}{2} = 0

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

- 1 + 2 sin (2 x) = 0

Move

1

to the right side

2 sin (2 x) = 1

Divide both sides by

2

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

General solutions for

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

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

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

sin (a) = \frac{2}{5}

2 cot^{2} (x) - 5 csc (x) - 1 = 0

3 sin (x) tan (x) + 3 tan (x) = 0

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

cos (A) = \frac{1}{3}

What is the general solution for 4sin(x)cos(x)-1=0 ?
The general solution for 4sin(x)cos(x)-1=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin