I need to solve for the angle, but I just don't know how to do it! The equation is: Cos A=Sin3A Can someone explain the formula so I can know how to do it, I will really appreciate it!

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  A = 22.5°, 45°, 112.5°, 202.5°, 225°, 292.5°

Step-by-step explanation:

Use the angle reduction formula to reduce this to an expression involving only sine or cosine, then solve the resulting higher-degree polynomial.

   sin(3x) = 3sin(x) -4sin³(x)

Using the identity for cosine, we have the radical equation ...

  √(1 -sin²(A)) = 3sin(A) -4sin³(A)

Squaring both sides, we have ...

  1 - sin²(A) = 9sin²(A) -24 sin⁴(A) +16sin⁶(A)

Letting z = sin²(A), this is the cubic equation ...

  1 -z = 9z -24z² +16z³

  16z³ -24z² +10z -1 = 0

Solving this by the usual methods, we find its factors to be ...

  (2z -1)(8z² -8z +1) = 0

  8(2z -1)((z -1/2)² -1/8) = 0 . . . . quadratic in vertex form

Then the roots are ...

  z = (1 ±√(1/2))/2, z = 1/2

The angle solutions are ...

  A = arcsin(±√z)

Of course, these need to be adjusted to fall in the domain of interest, and checked for extraneous solutions.

For z = 1/2-√(1/8), the angles are ...

  A = ±22.5°  ⇒  22.5°, 157.5°, 202.5°, 337.5°

For z = 1/2, the angles are ...

  A = ±45°  ⇒  45°, 135°, 225°, 315°

For z = 1/2+√(1/8), the angles are ...

  A = ±67.5°  ⇒  67.5°, 112.5°, 247.5°, 292.5°

The angles that are in bold are the actual solutions; the others are extraneous.

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All of the above is "the hard way." A much simpler approach is to take advantage of the identity ...

  cos(a) -sin(b) = (1/2)·sin(b/2 -a/2 -π/4)·sin(b/2 +a/2 -π/4)

Then you can use the zero product rule to find (for a=A, b=3A) ...

  (3A/2 -A/2 -π/4) = nπ  ⇒  A = nπ +π/4  ⇒  45°, 225°

  (3A/2 +A/2 -π/4) = nπ  ⇒  A = (nπ +π/4)  ⇒  22.5°, 112.5°, 202.5°, 292.5°

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Rewriting the equation as a function equal to zero, the solutions are the x-intercepts of the graph. In the attached, the x-axis is set to degrees.

  f(A) = cos(A) -sin(3A) = 0