Question

The number of solution of the solutions of the equation \[{\cot ^2}x - \cos e{c^{80}}x + 1 = 0\] in (0,13) is:

Answer 1

Hint: In order to solve this question we will first use the trigonometric identity then with the help of that identity we will transform this equation into a new equation then we will take the appropriate common from this equation. From this we will make two factors of that equation. And find the value of x from it.

Complete step by step solution:
For solving this question we will first take this equation and apply the trigonometric identity to transform it to new form:
Since the given equation was:
\[{\cot ^2}x - \cos e{c^{80}}x + 1 = 0\]
Now applying the trigonometric identity on it $ \left( {\cos e{c^2}\theta = 1 + {{\cot }^2}\theta } \right) $ we will get the new transformed equation:
 $ \cos e{c^2}x - \cos e{c^{80}}x = 0 $
Now taking one negative sign common from it we will get;
 $ \cos e{c^{80}}x - \cos e{c^2}x = 0 $
Now again taking $ \cos e{c^2}x $ as common from the above equation we will get;
 $ \cos e{c^2}x\left( {\cos e{c^{78}}x - 1} \right) = 0 $
As we can see that clearly that the value of $ \cos e{c^2}x $ can’t be 0 so the value of $ \left( {\cos e{c^{78}}x - 1} \right) $ will be equal to zero.
 $ \left( {\cos e{c^{78}}x - 1} \right) = 0 $
On substituting the values and on further solving this we will get;
 $ \cos e{c^{78}}x = 1 $
On converting it to sin term we will get:
 $ \dfrac{1}{{{{\sin }^{78}}x}} = 1 $
So we will get the value of:
 $ {\sin ^{78}}x = 1 $
Now we will notice that in (0,13) there are 4 possible values of x:
 $ \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2} $
Hence the number of solutions of x in (0,13) will be 4.
So, the correct answer is “4”.

Note: For solving this question we can also do one thing, that is we should also take one negative sign common and convert the whole equation in terms of cot and then we would have solved this whole equation in terms of cot.