What is the general solution for 2sin(x-10)=-0.4 ?

The general solution for 2sin(x-10)=-0.4 is x=-0.20135…+360n+10,x=180+0.20135…+360n+10

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2sin(x-10)=-0.4

2 sin (x - 1 0^{\circ}) = - 0.4

x = - 0.20135 \dots + 36 0^{\circ} n + 1 0^{\circ}, x = 18 0^{\circ} + 0.20135 \dots + 36 0^{\circ} n + 1 0^{\circ}

Solution steps

2 sin (x - 1 0^{\circ}) = - 0.4

Divide both sides by

2

sin (x - 1 0^{\circ}) = - 0.2

Apply trig inverse properties

x - 1 0^{\circ} = arcsin (- 0.2) + 36 0^{\circ} n, x - 1 0^{\circ} = 18 0^{\circ} + arcsin (0.2) + 36 0^{\circ} n

Solve

x - 1 0^{\circ} = arcsin (- 0.2) + 36 0^{\circ} n : x = - arcsin (\frac{1}{5}) + 36 0^{\circ} n + 1 0^{\circ}

Solve

x - 1 0^{\circ} = 18 0^{\circ} + arcsin (0.2) + 36 0^{\circ} n : x = 18 0^{\circ} + arcsin (0.2) + 36 0^{\circ} n + 1 0^{\circ}

x = - arcsin (\frac{1}{5}) + 36 0^{\circ} n + 1 0^{\circ}, x = 18 0^{\circ} + arcsin (0.2) + 36 0^{\circ} n + 1 0^{\circ}

Show solutions in decimal form

x = - 0.20135 \dots + 36 0^{\circ} n + 1 0^{\circ}, x = 18 0^{\circ} + 0.20135 \dots + 36 0^{\circ} n + 1 0^{\circ}

cot^{2} (2 x - \frac{π}{3}) = \frac{1}{3}

sin (2 x) = 0.8, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

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

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

\frac{12000}{2 \cdot 6380} = tan (\frac{9 0 ^{\circ} - u}{2})

What is the general solution for 2sin(x-10)=-0.4 ?
The general solution for 2sin(x-10)=-0.4 is x=-0.20135…+360n+10,x=180+0.20135…+360n+10