Verified answer

You can solve all of these the same way you would for an algebra problem. You try to isolate the variable (x). To do this, you apply inverse functions to both sides of the equation. For example, if you had 3x = 12, you would apply the inverse of multiplying by 3: divide by 3. So you'd get 3x/3 = 12/3 => x = 4.

These are essentially the same, except the functions are trigonometric. Fortunately, all of these functions have an inverse (well, not strictly) as well; for example, the inverse of cosine is usually written cos^-1 (that's cosine raised to the negative 1), or arccos.

Finding the inverse function is usually done by looking it up in a table or by using a calculator. If your teacher wants you to find it by hand, then I may not be able to help you since it has been a very long time since I've done that.

1) 3 cos x = 0.714

(3 cos x) / 3 = (0.714)/3 (divide both sides by 3)

cos x = 0.238

Now apply the inverse function:

arccos(cos(x)) = arccos(0.238)

x = arccos(0.238)

Using a calculator, we find that arccos of 0.238 is approximately 1.3305. Do keep in mind, however, that trigonometric functions are periodic, so there is actually more than 1 value of x that solves this equation. For example, 7.6137 also solves your equation, but your teacher probably wants you to find just one answer.

2) cos (x - 2) = -50

arccos(cos(x - 2)) = arccos(-50) [apply inverse]

x - 2 = arccos(-50)

x - 2 + 2 = arccos(-50) + 2 [add 2]

x = arccos(-50) + 2

x = 5.5746 + 2 = 7.5746

The other two are similar, so I'll omit the steps and show you what I got for an answer:

3) -1.4473

4) 0.0850