Question

How do you find all solutions to sin (2x + 1) = 0.2?

Answer 1

It has infinite set of solutions

Explanation:

The general solution is given as follows
`\sin(2x+1)=0.2`
`2x+1=2n\pi+\sin^{-1}(0.2)` OR `2x+1=2n\pi+\pi-\sin^{-1}(0.2)`
`x=\frac{2n\pi-1}{2}+\frac{1}{2}\sin^{-1}(0.2)` OR `x=\frac{(2n+1)\pi-1}{2}-\frac{1}{2}\sin^{-1}(0.2)`
where, `n` is any integer i.e. `n=0, \pm1,\pm2, \pm3, \ldots`

Answer 2

`x = 55^@57 + k360^@`
`x = 337^@11 + k360^@`

Explanation:

1 is expressed in radians. We can convert it to degrees for easier solving.
`pi = 3.14` --> `180^@`
1 radian --> `180/3.14 = 57^@32`
sin (2x + 57.32) = 0.2
Calculator and unit circle give 2 solutions for (2x + 57.32):
a. `(2x + 57.32) = 11^@54`
`2x = 11.54 - 57.32 = - 45^@78`
`x = - 22^@89`, or `x = 360 - 22.89 = 337^@11` (co-terminal).
b. `(2x + 57.32) = 180 - 11.54 = 168.46`
`2x = 168.46 - 57.32 = 111.14`
`x = 55^@57`
For general answers, add `k360^@`
Check by calculator.
`x = 55.57 --> 2x = 111.14 --> sin (2x + 57.32) = sin (168.46) = 0.20`. Proved