Question

··· Solve :-
sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x

!!...REQUIRED QUALITY ANSWER...!!
!!...Trigonometry...!!
​

Answer 1

We're asked to solve,

$\longrightarrow\sin x-3\sin(2x)+\sin(3x)=\cos x-3\cos(2x)+\cos(3x)$

or,

$\longrightarrow\sin(3x)+\sin x-3\sin(2x)=\cos(3x)+\cos x-3\cos(2x)$

We have,

• $\sin A+\sin B=2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$

• $\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$

Then,

$\footnotesize\longrightarrow2\sin\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right)-3\sin(2x)=2\cos\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right)-3\cos(2x)$

$\longrightarrow2\sin(2x)\cos x-3\sin(2x)=2\cos(2x)\cos x-3\cos(2x)$

Taking $\sin(2x)$ common in LHS and $\cos(2x)$ common in RHS,

$\longrightarrow\sin(2x)(2\cos x-3)=\cos(2x)(2\cos x-3)$

Taking RHS to LHS,

$\longrightarrow\sin(2x)(2\cos x-3)-\cos(2x)(2\cos x-3)=0$

$\longrightarrow(\sin(2x)-\cos(2x))(2\cos x-3)=0$

This implies,

$\longrightarrow2\cos x-3=0$

$\longrightarrow \cos x=\dfrac{3}{2}$

This is not possible since $\cos x\in\left[-1,\ 1\right]\not\cancel\ni\dfrac{3}{2}.$

And,

$\longrightarrow\sin(2x)-\cos(2x)=0$

$\longrightarrow\sin(2x)=\cos(2x)$

$\longrightarrow\dfrac{\sin(2x)}{\cos(2x)}=1$

$\longrightarrow\tan(2x)=1$

$\Longrightarrow 2x=n\pi+\dfrac{\pi}{4}$

$\longrightarrow\underline{\underline{x=\dfrac{n\pi}{2}+\dfrac{\pi}{8}}}$

This is the solution of the equation where $n\in\mathbb{Z}.$

Answer 2

ANSWER :

It ai given to solve the given pairs of Trigonometry, To solve,

Follow the steps...

⇒ 2 sin2x cosx − 3 sin2x − 2 cos2x cosx + 3 cos2x = 0

⇒ sin2x (2cosx − 3) − cos2x (2 cosx − 3) = 0

⇒ (sin2x − cos2x) (2 cosx − 3) = 0

⇒ sin2x = cos2x

⇒ 2x = 2nπ ± (π / 2 − 2x) i.e.,

x = nπ / 2 + π / 8

More Information :-

If f(x) and g(x) are basic trigonometric functions then period of [af (x) + bg(x)] = L.C.M.