How do u find all values of x in range -2(Pi) <= x <= 4(Pi) when :?

Question

a)sin(x) = 0.2317
b)cos(x) = -0.3154
c)sin(3x+1) = 0.7827
d)cos(2x-1.4) = -0.2167

Answer 1

Realize that what you are asking is what are the angles (x) in the previous rotation up through the next two rotations. Those are easy to find ONCE YOU HAVE THE SOLUTIONS IN THE FIRST ROTATION (i.e. from angle 0 to 2*pi). We'll focus on those first.

Each of these is solved by using inverse trig functions. For reference, arcsin and sin^(-1) are the same function.
a) x=arcsin(0.2317)
But, sine is positive in the 1st and 2nd quadrant, so once you get your answer, you'll use the reference angle to find the other reference angle in the 2nd quadrant.
b) x=arccos(-0.3154)
Cosine is negative in the 2nd and 3rd quadrant, so use the first quadrant reference angle to find the 2nd and 3rd quadrant solution.
c) 3x+1 = arcsin(0.7827)
x=(arcsin(0.7827)-1)/3
Same argument as before about sine being positive.
d) 2x-1.4=arccos(-0.2167) and then solve algebraically.
Same argument for part b)

Now, to take into account the answers in the other rotations, you'll need to add/subtract the proper amount to get into that rotation. For part a),
x=arcsin(0.2317) (whatever these angles in the 1st and 2nd quadrant are)
x=arcsin(0.2317) + 2*pi (these are the solutions in the second rotation, i.e. from 2*pi to 4*pi)
x=arcsin(0.2317) - 2*pi (will be the solutions in the previous rotation.

Repeat for parts b, c, & d).

Viola

Answer 2

the reply is e, more beneficial than 4. ln(x) receives infinitely detrimental close to 0, so subsequently it passes through a limiteless quantity of multiples of pi, and in view that sin(x) is 0 for multiples of pi (at the same time with detrimental multiples), there are a ton of zeroes.

Answer 3

a) x = 0.2338
b) x = 1.8917
c) 3x+1 = 0.8990
x = -0.0337
d) 2x-1.4 = 1.7892
x = 0.1946

All answers in radians, to 4 decimal places.