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

Question

Please show all the steps and formulas used

Answer 1

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, …

Answer 2

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.

 

Comments

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.

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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.

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See Fermat's answer.

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it is a very complicated question

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