Question

principe solution of coso=1-√3÷2√2
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:cos\theta = \dfrac{1 - \sqrt{3} }{2 \sqrt{2} }$

can be rewritten as

$\rm :\longmapsto\:cos\theta \: = - \: \bigg[ \dfrac{\sqrt{3} - 1}{2 \sqrt{2} } \bigg]$

Let first evaluate the value of

$\rm :\longmapsto\:\dfrac{ \sqrt{3} - 1}{2 \sqrt{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} }{2 \sqrt{2} } - \dfrac{1}{2 \sqrt{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{ \sqrt{2} } - \dfrac{1}{2} \times \dfrac{1}{ \sqrt{2} }$

$\rm \: = \: cos\bigg[\dfrac{\pi}{6} \bigg]cos\bigg[\dfrac{\pi}{4} \bigg] + sin\bigg[\dfrac{\pi}{6} \bigg]sin\bigg[\dfrac{\pi}{4} \bigg]$

We know,

$\boxed{ \tt{ \: cosx \: cosy \: - \: sinx \: siny \: = \: cos(x + y) \: }}$

So, using this identity, we get

$\rm \: = \: cos\bigg[\dfrac{\pi}{6} + \dfrac{\pi}{4} \bigg]$

$\rm \: = \: cos\bigg[\dfrac{2\pi + 3\pi}{12} \bigg]$

$\rm \: = \: cos\bigg[\dfrac{5\pi}{12} \bigg]$

So,

$\rm \implies\:\boxed{ \tt{\: cos\bigg[\dfrac{5\pi}{12} \bigg] = \frac{ \sqrt{3} - 1}{2 \sqrt{2} } \: }}$

Thus, the given Trigonometric expression can be rewritten as

$\rm :\longmapsto\:cos\theta \: = - \:cos \bigg[ \dfrac{5\pi}{12} \bigg]$

We know,

$\boxed{ \tt{ \: - cosx = cos(\pi - x) = cos(\pi + x) \: }}$

So, using this, the above can be rewritten as

$\rm :\longmapsto\:cos\theta \: = \:cos \bigg[\pi - \dfrac{5\pi}{12} \bigg] \: or \:cos \bigg[\pi + \dfrac{5\pi}{12} \bigg]$

$\rm :\longmapsto\:cos\theta \: = \:cos \bigg[\dfrac{12\pi - 5\pi}{12} \bigg] \: or \:cos \bigg[\dfrac{12\pi + 5\pi}{12} \bigg]$

$\rm :\longmapsto\:cos\theta \: = \:cos \bigg[\dfrac{7\pi}{12} \bigg] \: or \:cos \bigg[\dfrac{17\pi}{12} \bigg]$

$\bf\implies \:\theta = \dfrac{7\pi}{12} \: \: or \: \: \dfrac{17\pi}{12}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$