What is the general solution for 3sin(x+20)-0.75=0 ?

The general solution for 3sin(x+20)-0.75=0 is x=0.25268…+360n-20,x=180-0.25268…+360n-20

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3sin(x+20)-0.75=0

3 sin (x + 2 0^{\circ}) - 0.75 = 0

x = 0.25268 \dots + 36 0^{\circ} n - 2 0^{\circ}, x = 18 0^{\circ} - 0.25268 \dots + 36 0^{\circ} n - 2 0^{\circ}

Solution steps

3 sin (x + 2 0^{\circ}) - 0.75 = 0

Move

0.75

to the right side

3 sin (x + 2 0^{\circ}) = 0.75

Divide both sides by

3

sin (x + 2 0^{\circ}) = 0.25

Apply trig inverse properties

x + 2 0^{\circ} = arcsin (0.25) + 36 0^{\circ} n, x + 2 0^{\circ} = 18 0^{\circ} - arcsin (0.25) + 36 0^{\circ} n

Solve

x + 2 0^{\circ} = arcsin (0.25) + 36 0^{\circ} n : x = arcsin (0.25) + 36 0^{\circ} n - 2 0^{\circ}

Solve

x + 2 0^{\circ} = 18 0^{\circ} - arcsin (0.25) + 36 0^{\circ} n : x = 18 0^{\circ} - arcsin (0.25) + 36 0^{\circ} n - 2 0^{\circ}

x = arcsin (0.25) + 36 0^{\circ} n - 2 0^{\circ}, x = 18 0^{\circ} - arcsin (0.25) + 36 0^{\circ} n - 2 0^{\circ}

Show solutions in decimal form

x = 0.25268 \dots + 36 0^{\circ} n - 2 0^{\circ}, x = 18 0^{\circ} - 0.25268 \dots + 36 0^{\circ} n - 2 0^{\circ}

2 sin^{2} (x) + sin (x) - 5 = 0

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

\frac{3}{tan ( x )} = (2 tan (x)) (2 cos (x))

cos (θ) = - \frac{4}{5}, sin (θ)

5 sin (4 x) = 4 cos (2 x)

What is the general solution for 3sin(x+20)-0.75=0 ?
The general solution for 3sin(x+20)-0.75=0 is x=0.25268…+360n-20,x=180-0.25268…+360n-20