## Problem 19 – Solution

**The problem:**

**Solve the equation:**

*Admission exam to Mathematics,*

*Admission exam to Mathematics,*

*University of Bucharest, Romania, 19xx*

*University of Bucharest, Romania, 19xx*

**Our Solution:**

**Let’s start by calculating using formula **

**Where: and .**

**[1] **

**By replacing [1] in our equation we have:**

**[2] **

**From [2], we can see that the expression on the left side of the equation is when:**

**[3] or**

**[4] **

**a) Solving [3] we have: **** **

**We know that when**

**[5] in the first quadrant of the trigonometric circle.**

**[6] in the second quadrant of the trigonometric circle.**

** has a positive value therefore can only be in first and second quadrant.**

**If we consider angles larger than ( ), the general results for and are:**

**[7] and where is an integer.**

**b) Solving [4] we have: **** when:**

**[8] or in the first quadrant of the trigonometric circle. ( )**

**and**

**[9] or in the third quadrant of the trigonometric circle. ( )**

**If we consider angles larger than () and combine the two results, (for and from [8] and [9]), the general result for is:**

**[10] where is an integer and is the general form in which an odd number is shown.**

**The solutions for this equation are shown in [7] and [10].**

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