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

The general solution for sin(2x-10)=0.4 is x=180n+(0.41151…)/2+5,x=180n+90+5-(0.41151…)/2

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sin(2x-10)=0.4

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

x = 18 0^{\circ} n + \frac{0.41151 \dots}{2} + 5^{\circ}, x = 18 0^{\circ} n + 9 0^{\circ} + 5^{\circ} - \frac{0.41151 \dots}{2}

Solution steps

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

Apply trig inverse properties

2 x - 1 0^{\circ} = arcsin (0.4) + 36 0^{\circ} n, 2 x - 1 0^{\circ} = 18 0^{\circ} - arcsin (0.4) + 36 0^{\circ} n

Solve

2 x - 1 0^{\circ} = arcsin (0.4) + 36 0^{\circ} n : x = 18 0^{\circ} n + \frac{arcsin ( 0.4 )}{2} + 5^{\circ}

Solve

2 x - 1 0^{\circ} = 18 0^{\circ} - arcsin (0.4) + 36 0^{\circ} n : x = 18 0^{\circ} n + 9 0^{\circ} + 5^{\circ} - \frac{arcsin ( 0.4 )}{2}

x = 18 0^{\circ} n + \frac{arcsin ( 0.4 )}{2} + 5^{\circ}, x = 18 0^{\circ} n + 9 0^{\circ} + 5^{\circ} - \frac{arcsin ( 0.4 )}{2}

Show solutions in decimal form

x = 18 0^{\circ} n + \frac{0.41151 \dots}{2} + 5^{\circ}, x = 18 0^{\circ} n + 9 0^{\circ} + 5^{\circ} - \frac{0.41151 \dots}{2}

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

- 6 cos^{2} (θ) - 9 cos (θ) - 7 = 8 cos (θ)

θ, 1 + cos (θ) = 1 - sin (θ)

sin (x) = 1

2 tan^{2} (θ) - tan (θ) = 0

What is the general solution for sin(2x-10)=0.4 ?
The general solution for sin(2x-10)=0.4 is x=180n+(0.41151…)/2+5,x=180n+90+5-(0.41151…)/2