Question

How do you solve `sin (cosx) = sqrt2/4`?

Answer 1

Take the arcsin to determine cos(x); then take arccos to determine `x` (= 1.201)

Explanation:

If `sin(cos(x)) = sqrt(2)/4`
then
`color(white)("XXXX")` `cos(x)`
`color(white)("XXXX")` `= arcsin(sin(cos(x)))`

`color(white)("XXXX")` `=arcsin(sqrt(2)/4)`
`color(white)("XXXX")` `color(white)("XXXX")`(using a calculator)
`color(white)("XXXX")` `=0.361367`

If `cos(x) = 0.361367`
then
`color(white)("XXXX")` `x`
`color(white)("XXXX")` `= arccos(cos(x))`

`color(white)("XXXX")` `=arccos(0.361367)`
`color(white)("XXXX")` `color(white)("XXXX")`(again, using a calculato)
`color(white)("XXXX")` `= 1.201`

Note: I have assumed all values are in radians

Answer 2

Solve `sin (arccos (sqrt2/4))`

Explanation:

On the trig unit circle,

`cos x= sqrt2/4 = 0.35` --> `x = +- 69.30` deg

a . sin y = sin (69.30) = 0.94-->

y = 69.30 and y = 180 - 69.30 = 110.70 deg

b. sin y = sin (-69.30) = -0.94 -> sin (360 - 69.30) = 290.70

and y = 180 + 69.30 = 249.30

Within period (0, 360), there are 4 answers;

69.30; 110.70; 249.30 and 290.70.

Check by calculator: cos 290.70 = cos 249.30 = cos 110.70 = cos 69.30 = 0.35

Answer 3

Alternate Answer The given equation can not be solved since `sin(cos(x))` is a meaningless expression.

Explanation:

`cos x` can be interpreted to mean one of two things:
1. `cos_r(x)` where `x` is an angle measured in radians;
2. `cos_d(x)` where `x` is an angle measured in degrees.

Whichever interpretation is used
`cos x` is a ratio of tow sides of a right-triangle (it is not an angle measurement).

Similarly, `sin(theta)` can be interpreted as
1. `sin_r(theta)` where `theta` is an angle measured in radians;
2. `sin_d(theta)` where `theta` is an angle measured in degrees.

Note that `sin(theta)` does not have any meaning unless `theta` is an angle.
In particular `sin(theta)` is meaningless if `theta` is not an angle.

Since the value of `cos x` is not an angle,
`sin(cos x)` is meaningless.