Solve for x - sin3- - 4sin-sinx - - sinx - - where - is a constant - n- - n- I-

The correct option is A nπ±π/3, n∈ I #160;sin 3α #160; =4sinα #160;sin( x+α #160;) #160;sin( x α #160;) #160; #160;⇒sin 3α #160 ...

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

Byju's Answer

Standard XII

Mathematics

Property 2

Solve for x :...

Question

Solve for x : $sin 3 α = 4 sin α sin (x + α) sin (x - α)$ where $α$ is a constant $\neq n π, n \in I$ .

n π \pm \frac{π}{3}, n \in I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n π - \frac{π}{3}, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n π + \frac{π}{3}, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n π \pm \frac{π}{6}, n \in I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $n π \pm \frac{π}{3}, n \in I$
$sin 3 α = 4 sin α sin (x + α) sin (x - α)$
$\Rightarrow sin 3 α = 4 sin α (sin x cos α + cos x sin α) \times (sin x cos α - cos x sin α)$
$\Rightarrow sin 3 α = 4 sin α ({sin}^{2} x {cos}^{2} α - {cos}^{2} x {sin}^{2} α)$
$\Rightarrow sin 3 α = 4 sin α ({sin}^{2} x (1 - {sin}^{2} α) - (1 - {sin}^{2} x) {sin}^{2} α)$
$\Rightarrow sin 3 α = 4 sin α ({sin}^{2} x - {sin}^{2} x {sin}^{2} α - {sin}^{2} α + {sin}^{2} x {sin}^{2} α)$
$\Rightarrow 3 sin α - 4 {sin}^{3} α = 4 sin α ({sin}^{2} x - {sin}^{2} α)$
$\Rightarrow 3 sin α - 4 {sin}^{3} α = 4 sin α {sin}^{2} x - 4 {sin}^{3} α$
$\Rightarrow 3 sin α = 4 sin α {sin}^{2} x$
$\Rightarrow {sin}^{2} x = \frac{3}{4} \Rightarrow x = n π \pm \frac{π}{3}$

Suggest Corrections

Similar questions

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

$\sqrt{3} + i = (a + i b) (c + i d), t h e n {tan}^{- 1} \frac{b}{a} + {tan}^{- 1} \frac{d}{c}$

has the value

Q. If

sin 3 α = 4 sin α sin (x + α) sin (x - α)

and

α \neq n π, n \in I

then

864 {sin}^{2} x + 3620 {cos}^{2} x

is equal to

The value of x $\in (- 2 π, 2 π)$ such that $\frac{s i n x + i c o s x}{1 + i}$ , where $i = \sqrt{- 1}$ , is purely imaginary are given by

Q. General solution of

x

for which

{cos}^{2} x = \frac{sin x tan x}{6}

is :

Join BYJU'S Learning Program

Solve for x : sin3α=4sinαsin(x+α)sin(x−α) where α is a constant ≠nπ,n∈I.

Solve for x : $sin 3 α = 4 sin α sin (x + α) sin (x - α)$ where $α$ is a constant $\neq n π, n \in I$ .