What is the general solution for sin(x)= 1/5 ,cos(x)>0 ?

The general solution for sin(x)= 1/5 ,cos(x)>0 is No Solution for x\in\mathbb{R}

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sin(x)= 1/5 ,cos(x)>0

sin (x) = \frac{1}{5}, cos (x) > 0

No Solution for x \in R

Solution steps

sin (x) = \frac{1}{5}, cos (x) > 0

Apply trig inverse properties

sin (x) = \frac{1}{5}

General solutions for

sin (x) = \frac{1}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

Solutions for the range

cos (x) > 0

No Solution for x \in R

2 sin^{2} (2 θ) = - 1 - 3 sin (2 θ)

6 cos (x) = 3

5 cot (x) = cos (x)

2 sin^{2} (x) - 1 = cos (x)

9 sin^{2} (θ) + 19 sin (θ) + 8 = 7 sin (θ) + 4

What is the general solution for sin(x)= 1/5 ,cos(x)>0 ?
The general solution for sin(x)= 1/5 ,cos(x)>0 is No Solution for x\in\mathbb{R}