Question

The number of values of $x\in [0,2\pi ]$ satisfying the equation $\sin 2x+5\sin x+1+5\cos x=0$ is

• A. 02
• B. 2.0
• C. 2

Answer 1

Answer: (A), (B), (C)

Given that
$\sin 2x+5\sin x+1+5\cos x=0$
When we have trigonometric equations involving $(\sin x\pm \cos x)$$\&\sin x\cos x\text{or}\sin 2x,$
We can solve it by the substitution $\sin x\pm \cos x=t$.

Now, if we observe the given expression, we can re-write it as
$1+\sin 2x+5(\sin x+\cos x)$
This is an expression involving $\sin x+\cos x\text{and}\sin 2x.$
So we will go for the substitution $\sin x+\cos x=t.$

It is important to know that $\sin x+\cos x\text{and}\sin 2x$ are connected by the relation:
$(\sin x+\cos x{)}^2=1+\sin 2x.[$we also have $(\sin x-\cos x{)}^2=1-\sin 2x]$
$\Rightarrow 1+\sin 2x+5(\sin x+\cos x)=t^2+5t.$

$\Rightarrow t^2+5t=0$

$\Rightarrow t=0\text{or}t+5=0$

$\Rightarrow \sin x+\cos x=0\text{or}\sin x+\cos x+5=0$

$\sin x+\cos x+5\ne 0\forall x\in R$
$\Rightarrow \sin x+\cos x=0$

$\Rightarrow \tan x=-1=\tan \frac{-\pi }{4}$

$\Rightarrow x=n\pi -\frac{\pi }{4}\forall n\in Z$

$x=\frac{-\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4},\frac{15\pi }{4}......$ out of which $\frac{3\pi }{4},\frac{7\pi }{4}\in [0,2\pi ].$
Therefore, there are only two values in the given interval.