What is the general solution for 4sin(2x)-3=0 ?

The general solution for 4sin(2x)-3=0 is x=(0.84806…)/2+pin,x= pi/2-(0.84806…)/2+pin

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4sin(2x)-3=0

4 sin (2 x) - 3 = 0

x = \frac{0.84806 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.84806 \dots}{2} + πn

Solution steps

4 sin (2 x) - 3 = 0

Move

3

to the right side

4 sin (2 x) = 3

Divide both sides by

4

sin (2 x) = \frac{3}{4}

Apply trig inverse properties

2 x = arcsin (\frac{3}{4}) + 2 πn, 2 x = π - arcsin (\frac{3}{4}) + 2 πn

Solve

2 x = arcsin (\frac{3}{4}) + 2 πn : x = \frac{arcsin ( \frac{3}{4} )}{2} + πn

Solve

2 x = π - arcsin (\frac{3}{4}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{3}{4} )}{2} + πn

x = \frac{arcsin ( \frac{3}{4} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{3}{4} )}{2} + πn

Show solutions in decimal form

x = \frac{0.84806 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.84806 \dots}{2} + πn

cos (a) = \frac{5}{13}, e = 5 \cdot (tan (a) + sec (a))

csc (\frac{x}{2}) - 1 = 0

4 cos (x) + 4 sin (x) = 26

7 tan^{2} (θ) - 4 tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

3 - 4 sin^{2} (x) = 2 cos^{2} (x)

What is the general solution for 4sin(2x)-3=0 ?
The general solution for 4sin(2x)-3=0 is x=(0.84806…)/2+pin,x= pi/2-(0.84806…)/2+pin