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

Last updated date: 21st Jun 2024
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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.