If sin 2x - 1- then 0 cos x -sin x sin x 0 cos x cos x sin x 0 equals -

sin 2x =1 ⇒ 2x=π #160; 2 x=π #160; 4 0 amp; cos x #160; amp; sin x #160; sin x #160; amp; 0 amp; cos x #160 ...

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Definition of a Determinant

If sin 2x =...

Question

If $s i n 2 x = 1,$ then $∣ ∣ ∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣ ∣ ∣$ equals .............

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S = ⎧ ⎪ ⎨ ⎪ ⎩ x ϵ [0, 2 π] : ∣ ∣ ∣ ∣ \begin{matrix} 0 & cos x & - sin x sin x & 0 & cos x cos x & sin x & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0 ⎫ ⎪ ⎬ ⎪ ⎭

\sum x ϵ S tan (\frac{π}{3} + x)

f (x) = {∣ ∣ ∣ ∣ \begin{matrix} 0 & cos x & - sin x sin x & 0 & cos x cos x & sin x & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2}

sin 2 x = 1

f (x) = 2 / 3

f (x) = 0

sin x = cos x

{∣ ∣ ∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2}

sin 2 x = α - 1

cos 2 x = β - 1

\frac{{sec}^{2} x [({cos}^{2} x - {sin}^{2} x) - 2 sin x cos x]}{1 + sin 2 x}

f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})

g (\frac{5}{4}) = 1,

g o f (x)

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