Question

Find the general solution: \[{\cot ^2}x + \dfrac{3}{{\sin x}} + 3 = 0\]

Answer 1

Hint: we know the solutions where \[0 \leqslant x \leqslant 2\pi \]
General solution: the expression involving integer ‘n’ this gives all solutions of trigonometric equation, here we use general formula.
\[if\sin \theta = 0,\]then\[\theta = n\pi \]
\[if\sin x = \sin y\]then\[x = n\pi + ( - 1)ny,\]where \[n \in Z\]

Complete step by step answer:

\[{\cot ^2}x + \dfrac{3}{{\sin x}} + 3 = 0\]
\[\cos e{c^2}x - 1 + 3\cos ecx + 3 = 0\]
\[\cos e{c^2}x + 3\cos ecx + 2 = 0\]
\[\cos e{c^2}x + 2\cos ecx + \cos ecx + 2 = 0\]
\[(\cos ecx + 2) = 0\]and\[(\cos ecx + 1) = 0\]
\[\cos ecx = - 2\]
Now \[\dfrac{1}{{\sin x}} = - 2\]
\[\sin x = - \dfrac{1}{2}\]
We know \[\sin 30^\circ = \dfrac{1}{2}\]
\[\sin \theta \]is negative in III and IV quadrants
\[\sin x = \sin (\pi + 30^\circ )\]
\[\sin x = \sin 210^\circ \]
\[x = 210^\circ \]or\[(\pi + \dfrac{\pi }{6}) = \dfrac{{7\pi }}{6}\]
 Now \[\sin x = \sin \dfrac{{7\pi }}{6}\]
General solution = \[n\pi + {( - 1)^n}\dfrac{{7\pi }}{6}\]
Now \[\cos ecx = - 1\]
\[\dfrac{1}{{\sin x}} = - 1\]
\[\sin x = - 1\]
We know \[\sin 90^\circ = 1\]
\[\sin \theta \]is negative in III and IV quadrants
\[\sin x = \sin (\pi + \dfrac{\pi }{2})\]
\[\sin x = \sin \dfrac{{3\pi }}{2}\]
General solution = \[n\pi + {( - 1)^n}\sin \dfrac{{3\pi }}{2}\]

Note: The equations that involve the trigonometric functions of a variable are called trigonometric equations These equations have one or more trigonometric ratios of unknown angles. For example,\[cos{\text{ }}x - si{n^2}\;x = 0\] is a trigonometric equation which does not satisfy all the values of \[x\]. Hence for such equations, we have to find the values of \[x\] or find the solution.
We know that \[sin{\text{ }}x\]and \[cos{\text{ }}x\] repeat themselves after an interval of\[2\pi \], and \[tan{\text{ }}x\]repeats itself after an interval of\[\pi \]. The solutions such trigonometry equations which lie in the interval of \[\left[ {0,{\text{ }}2\pi } \right]\] are called principal solutions. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation, and it is expressed in a generalized form in terms of ‘n’.
we can also find the solution by taking values in the fourth quadrant. And we can find a general solution by knowing the range of that trigonometric function.