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

2 Answers

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

Jun 19, 2018

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