Question

See the question of trignometryin photo and please

it ​

Answer 1

I am not a jee student. but whatever I know I did . hope it helps.

Answer 2

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▪Answer :-

$\bf \large \green{2\pi}$

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▪Given :-

S = { θ ∈ [-2π,2π] : 2cos²θ + 3sinθ = 0 }.

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▪To Calculate :-

Sum of elements of S.

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▪Solution :-

$\sf \: We \: have \: \: \theta \in[ - 2\pi,2\pi] \\ \\ \sf and \\ \\ \sf 2 {cos}^{2} \theta + 3sin\theta = 0 \\ \\ \sf\implies 2(1 - {sin}^{2} \theta )+ 3sin\theta = 0 \\ \\ \implies \sf 2 - 2 {sin}^{2} \theta + 3sin\theta = 0 \\ \\ \bf \implies2 {sin}^{2} \theta - 3sin\theta - 2 = 0$

Which is quadratic equation in sinθ.

《Solving the equation》

$\sf \implies2 {sin}^{2} \theta - 3sin\theta - 2 = 0 \\ \\ \implies \sf 2 {sin}^{2}\theta - 4sin \theta + sin\theta - = 0\\ \\ \implies \sf 2sin \theta(sin \theta - 2) + 1(sin \theta - 2)\\ \\ \implies \sf(2sin \theta + 1)(sin\theta - 2) = 0 \\ \\ \implies \colorbox{lime}{ \bf \underline{ \boxed{ \bf sin\theta = - \frac{1}{2} }}} \: \: \: \: \{ \bf\because sin\theta \ne2 \}$

As θ ∈ [-2π,2π]

So,

$\sf\theta = 2\pi - \frac{\pi}{6} \\ or \\ \sf \theta = - \pi + \frac{\pi}{6} \\ or \\ \sf\theta = \frac{ - \pi}{6 } \\ or \\ \sf\theta = \sf\pi + \frac{\pi}{6}$

Hence Sum of All Solutions :-

$\sf 2\pi - \frac{\pi}{6} + - \pi + \frac{\pi}{6} + \frac{ - \pi}{6 } + \sf\pi + \frac{\pi}{6} \\ \\ \huge \green{ = 2\pi}$

$\red{\Large \mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required }} \\ \huge \red{ \mathfrak{ \text{ A}nswer.}}$

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