Solve- 1-sin 2 - cos 2- 4sin 4- sin 2 - 1-cos 2 - 4sin 4- sin 2 - cos 2 - 1-4sin 4- -0- where- 0- -2-

Given #160;1+sin ^2 θ amp; cos ^2θ amp; 4sin 4θ #160;sin ^2 θ amp;1+cos ^2 θ #160; amp; 4sin 4θ #160;sin ^2 θ amp; cos ^2 θ #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Determinant

Solve: 1+si...

Question

Solve:
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= 0$ ; where, $0 < θ < \frac{π}{2}$ .

Open in App

Solution

Given $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= 0$ ; where, $0 < θ < \frac{π}{2}$ .
$(C_{1} \to C_{1} + C 2)$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & {cos}^{2} θ & 4 sin 4 θ 2 & 1 + {cos}^{2} θ & 4 sin 4 θ 1 & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= 0$

$(R_{2} \to R_{2} - R_{1}$ & $R_{3} \to R_{3} - R_{1})$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 2 & {cos}^{2} θ & 4 sin 4 θ 0 & 1 & 0 - 1 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$ $= 0$
$\Rightarrow 2 + 4 sin 4 θ = 0$

$\Rightarrow sin 4 θ = - \frac{1}{2}$

$\Rightarrow 4 θ = \frac{7 π}{6}$ ...[Since $0 < θ < \frac{π}{2}$ ]
$\Rightarrow θ = \frac{7 π}{24}$

Suggest Corrections

Similar questions

If $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ such that $0 \leq θ \leq \frac{π}{2}$ then $θ$ is

Q. Solve

5 cos 2 θ + 2 {cos}^{2} \frac{θ}{2} + 1 = 0, - π < 0 < π

Q. Solve:

5 c o s 2 θ + 2 c o s^{2} \frac{θ}{2} + 1 = 0, - π < θ < π

The value/s of $θ$ lying between 0 and $\frac{π}{2}$ and satisfying the equation $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0, a r e$

Q. Solve:

5 cos 2 θ + 2 {cos}^{2} (\frac{1}{2} θ) + 1 = 0, - π < θ < π

Join BYJU'S Learning Program

Solve: ∣∣ ∣ ∣∣1+sin2θcos2θ4sin4θsin2θ1+cos2θ4sin4θsin2θcos2θ1+4sin4θ∣∣ ∣ ∣∣ =0; where, 0<θ<π2.

Solve:
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= 0$ ; where, $0 < θ < \frac{π}{2}$ .