Question

Solve theta : 2sin theta cos theta = 2 − sintheta + 4costheta , (−π≤ theta ≤ π).

Maths Aryabhatta

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Answer 1

Answer:

Here's your answer

Step-by-step explanation:

∫

(4+cos

2

θ)(2−sin

2

θ)

sinθ

dθ

Lety x=cosθ⇒dx=−sinθdθ

−∫

(4+x

2

)(1+x

2

)

dx

∣∴2sin

2

θ=1+(1−sin

2

θ)=1+cos

2

θ=1+x

2

=−

3

1

∫(

1+x

2

1

−

4+x

2

1

)dx

=

3

−1

[tan

−1

x−

2

1

tan

−1

(

2

x

)]

=

3

1

[tan

−1

(cosθ)−

2

1

tan

−1

(

2

cosθ

)]

Hope it helps

Pls mark me the brainliest

Answer 2

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

$\rm :\longmapsto\: - \pi \leqslant \theta \leqslant \pi$

and

$\rm :\longmapsto\:2sin \theta cos\theta \: = \: 2 - sin\theta + 4cos\theta$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \rm{ \theta }}$

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

Given Trigonometric equation is

$\rm :\longmapsto\:2sin \theta cos\theta \: = \: 2 - sin\theta + 4cos\theta$

can be rewritten as

$\rm :\longmapsto\:2sin \theta cos\theta - \: 2 + sin\theta - 4cos\theta = 0$

can be re-arranged as

$\rm :\longmapsto\:2sin \theta cos\theta + sin\theta - 2 - 4cos\theta = 0$

$\rm :\longmapsto\:sin\theta (2cos\theta + 1) - 2(1 + 2cos\theta ) = 0$

$\rm :\longmapsto\:(sin\theta - 2)(2cos\theta + 1) = 0$

$\rm :\implies\:sin\theta = 2 \: \: \: or \: \: \: cos\theta = - \dfrac{1}{2}$

Now,

$\red{\rm :\implies\:sin\theta = 2 \: is \: rejected \: as \: - 1 \leqslant sin\theta \leqslant 1}$

So,

$\rm :\implies\:cos\theta = - \: \dfrac{1}{2}$

$\rm :\implies\:cos\theta = \: cos\dfrac{2\pi}{3}$

We know,

$\boxed{ \rm{ cosx = cosy \: \implies \: x = 2n\pi \: \pm \: y \: \forall \: n \: \in \: Z}}$

So,

$\bf\implies \:\theta = 2n\pi \: \pm \: \dfrac{2\pi}{3} \: \forall \: n \: \in \: Z$

As it is given that,

$\rm :\longmapsto\: - \pi \leqslant \theta \leqslant \pi$

So, it can assume the values,

$\bf\implies \:\theta \: = \: \dfrac{2\pi}{3}, \: - \: \dfrac{2\pi}{3}$

Additional Information : -

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