Question

if y = x sin x^2 then the value of dy ?dx is​

Answer 1

Answer:

Dy/dx=xcosx^2(2x)+sinx^2

=2x^2cosx^2+sinx^2

Answer 2

Answer:

Answer

The general solution of the given equation is :

2nπ ± π/3 , n € Z

Given

The trigonometric equation is

\bullet \: \: \sf \sin^{2}\theta - 2\cos\theta + \dfrac{1}{4} = 0∙sin

2

θ−2cosθ+

4

1

=0

To Find

The general value of \sf \thetaθ

Solution

Given ,

\begin{gathered} \sf \sin {}^{2} \theta - 2 \cos \theta + \dfrac{1}{4} = 0 \\ \\ \sf\implies1 - \cos {}^{2} \theta - 2 \cos \theta + \dfrac{1}{4} = 0 \\ \\ \sf \implies4 - 4 \cos {}^{2} \theta - 8 \cos \theta + 1 = 0 \\ \\ \sf \implies - 4 \cos {}^{2} \theta - 8 \cos \theta + 5 = 0 \\ \\ \sf \implies4 \cos {}^{2} \theta + 8\cos \theta - 5 = 0 \\ \\ \sf \sf\implies4 { \cos}^{2} \theta - 2 \cos \theta + 10 \cos \theta - 5 = 0 \\ \\ \sf \implies2 \cos \theta( \cos \theta - 1) + 5(2 \cos \theta - 1) = 0 \\ \\ \sf \implies( 2\cos \theta - 1)(2 \cos \theta + 5) = 0\end{gathered}

sin

2

θ−2cosθ+

4

1

=0

⟹1−cos

2

θ−2cosθ+

4

1

=0

⟹4−4cos

2

θ−8cosθ+1=0

⟹−4cos

2

θ−8cosθ+5=0

⟹4cos

2

θ+8cosθ−5=0

⟹4cos

2

θ−2cosθ+10cosθ−5=0

⟹2cosθ(cosθ−1)+5(2cosθ−1)=0

⟹(2cosθ−1)(2cosθ+5)=0

Now we have ,

\begin{gathered}\sf\implies 2\cos\theta - 1 = 0 \: \: and \: \: implies 2\cos\theta + 5 = 0 \\\\ \sf\implies \cos\theta = \dfrac{1}{2} \: \: and \: \implies \cos\theta = \dfrac{-5}{2} (not \: possible ) \end{gathered}

⟹2cosθ−1=0andimplies2cosθ+5=0

⟹cosθ=

2

1

and⟹cosθ=

2

−5

(notpossible)

So we are taking ,

\begin{gathered}\sf\implies \cos\theta = \dfrac{1}{2} \\\\ \sf\implies \cos\theta = \cos\dfrac{\pi}{3} \\\\ \sf\implies x = 2n\pi \pm \dfrac{\pi}{3} , n \in Z \end{gathered}

⟹cosθ=

2

1

⟹cosθ=cos

3

π

⟹x=2nπ±

3

π

,n∈Z