Question

Solve my question mods​

Answer 1

Answer:

Given,

• sin⁴x + cos⁴x = sin x.cos x   ...(i)

We know that,

⇒ (sin²x + cos²x)² = sin⁴x + cos⁴x + 2.sin²x. cos²x

⇒ 1 = sin⁴x + cos⁴x + 2.sin²x. cos²x

⇒ 1 - 2.sin²x. cos²x = sin⁴x + cos⁴x  ...(ii)

Substituting the LHS of (ii) in the LHS of (i) we get:

⇒ 1 - 2.sin²x. cos²x = sin x.cos x

⇒ 1 - 2.sin²x. cos²x - sin x.cos x = 0

Multiplying the whole equation by 2, we get:

⇒ 2 - 4.sin²x. cos²x - 2.sin x.cos x = 0

⇒ 2 - (2.sin x.cos x)² - 2.sin x.cos x = 0

We know that, 2.sin x.cos x = sin 2x. Hence substituting it, we get:

⇒ 2 - (sin 2x)² - sin 2x = 0

Let sin 2x = t. Hence we get:

⇒ 2 - t² - t = 0

⇒ t² + t - 2 = 0

⇒ t² + 2t - t - 2 = 0

⇒ t ( t + 2 ) - 1 ( t + 2 ) = 0

⇒ ( t + 2 ) ( t - 1 ) = 0

⇒ t = -2 and t = 1

Sin function cannot have a value of -2. Hence we consider only t = 1.

Substituting t back to sin 2x,

⇒ sin 2x = 1

We know that, sin takes the value of 1 at π/2, 5π/2, 7π/2, etc. Hence the general form is given as:

⇒ 2x = 2nπ + π/2

⇒ x = nπ + π/4

Hence Option (A) is the correct answer.

Answer 2

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

Given Trigonometric equation is

$\rm :\longmapsto\: {sin}^{4}x + {cos}^{4}x = sinx \: cosx$

can be rewritten as

$\rm :\longmapsto\: {( {sin}^{2}x) }^{2} + {( {cos}^{2}x) }^{2} = sinxcosx$

$\rm :\longmapsto\: {( {sin}^{2}x) }^{2} + {( {cos}^{2}x) }^{2} + 2 {sin}^{2}x{cos}^{2}x - 2 {sin}^{2}x{cos}^{2}x = sinxcosx$

$\rm :\longmapsto\: {( {sin}^{2}x + {cos}^{2}x )}^{2} - 2 {sin}^{2}x{cos}^{2}x = sinx \: cosx$

$\rm :\longmapsto\: {1}^{2} - 2 {sin}^{2}x{cos}^{2}x = sinx \: cosx$

$\rm :\longmapsto\: 1 - 2 {sin}^{2}x{cos}^{2}x = sinx \: cosx$

Multiply whole equation by 2, we get

$\rm :\longmapsto\: 2 - 4{sin}^{2}x{cos}^{2}x = 2sinx \: cosx$

$\rm :\longmapsto\: 2 - {(2sinx \: cosx)}^{2} = 2sinx \: cosx$

We know,

$\boxed{ \bf{ \:2sinxcosx = sin2x}}$

So, using this identity, we get

$\rm :\longmapsto\:2 - {sin}^{2}2x = sin2x$

can be rewritten as

$\rm :\longmapsto\: {sin}^{2}2x + sin2x - 2 = 0$

$\rm :\longmapsto\: {sin}^{2}2x + 2sin2x - sin2x - 2 = 0$

$\rm :\longmapsto\:sin2x(sin2x + 2) - 1(sin2x + 2) = 0$

$\rm :\longmapsto\:(sin2x + 2)(sin2x - 1) = 0$

$\bf\implies \:sin2x = 1 \: \: and \: \: sin2x = - 2 \: \{rejected \}$

$\bf\implies \:sin2x = sin\dfrac{\pi}{2}$

$\bf\implies \:2x = n\pi \: + \: {( - 1)}^{n} \: \dfrac{\pi}{2} \: \forall \: n \in \: Z$

$\bf\implies \:x = \dfrac{n\pi}{2} \: + \: {( - 1)}^{n} \: \dfrac{\pi}{4}$

$\bf\implies \:x = \dfrac{\pi}{4},\dfrac{5\pi}{4} ,\dfrac{9\pi}{4} , - - -$

$\bf\implies \:x = n\pi + \dfrac{\pi}{4} \: \: \forall \: n \in \: Z$

• Hence, option (A) is correct.

Additional Information :-

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