Please show all the steps and formulas used

Cot2x - tan 78 = ( sec x sec 78 )/2

​Here's another solution

I showed this to Rod and he corrected an error in the final steps, The full solution is as follows.

I let y = 78 degrees in the expression above and simplified it from there.

cot(2x) – tan(y) = ( sec(x).sec(y) )/2

cos(2x)/sin(2x) – sin(y)/cos(y) = 1/(2cos(x).cos(y))

{cos(2x).cos(y) – sin(2x).sin(y)}/{sin(2x).cos(y)} = 1/(2cos(x).cos(y))

{cos(2x+y)}/{sin(2x).cos(y)} = 1/(2cos(x).cos(y))

{cos(2x+y)}/{sin(2x)} = 1/(2cos(x))

cos(2x+y) = sin(2x)/(2cos(x))

cos(2x+y) = 2sin(x).cos(x)/(2cos(x))

cos(2x+y) = sin(x)

cos(2x+y)  = cos(pi/2-x)

equating the arguments of the two cosines,

2x + y = pi/2 – x + 2n.pi

3x = pi/2 + 2n.pi – y

3x = 90 + 360n – 78

3x = 12 + 360n

​x  = 4 + 120n

x = 4 + 0, 120, 240, 360, …

by Level 11 User (80.9k points)
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Solve for x: x=4 is the solution. See details below.

Here's how I actually solved it using a calculator: I made a table using the formula as above. Then I looked for a cell containing 0.0000. The X value for this row was the solution, so x=4. But because this seemed so exact I wanted to find a solution based on trigonometry.

Another approach is to realise that sec(x)sec(78)/2 is pretty constant for small values of x=sec(78)/2=2.4 approximately (see blue graph below). So tan(90-x)=tan(78)+2.4=7.1 approx. So 90-x=arctan(7.1)=81.98 approx, making 2x=8.02 and x=4 approx. Then substituting x=4 in the original equation we find we have the answer because the equation balances! The graphs below show the functions: red shows the left-hand side function, blue the right-hand side and green is left-hand side minus right-hand side. At x=4 the left and right sides are equal. Please see Fermat's solution for more solutions for x.

by Top Rated User (711k points)

Hi, Mac2016.

I'm still trying to find out why x=4 degrees. Thank you for marking this as Best Answer, but I'm still working on finding the easiest way to find x. So far I have only found complicated ways, and I strongly feel there's an easier solution. If and when I find one I will amend my answer accordingly.

A fellow user has been in touch with me with a proper solution to your problem. I hope he's going to post his solution soon. His will then be the best answer. The graph and calculator solutions I presented are still valid but I was unable to find the trigonometrical solution.

it is a very complicated question

I went about it the wrong way. Fermat's solution is much more straightforward.