Question

How many values of $xϵ[0,2\pi ]$ satisfies the equation sin 2x + 5 sin x + 1 + 5 cos x = 0?

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Answer 1

When we have trigonometric equations involving $(sinx\pm cosx)$ and sin x cos x or sin 2x, we can solve it

by the substitution $sinx\pm cosx=t$.
If we observe the expression given, we can re-write it as

1+ sin 2x + 5(sin x + cos x)
This is an expression involving sin x + cos x 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 and sin 2x are connected by the relation

$(sinx+cosx{)}^2$ =1+ sin 2x. [we also have $(sinx-cosx{)}^2$=1-sin 2x]
$\Rightarrow$ 1 + sin 2x + 5(sin x + cos x) = $t^2$ + 5t.

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

$\Rightarrow$ t=0 or t+5=0

$\Rightarrow$ sin x + cos x = 0 or sin x + cos x + 5 = 0

sin x + cos x + 5 can't be zero, because minimum value of sin x + cos x is -$\sqrt{2}$

$\Rightarrow$ sin x + cos x = 0

$\Rightarrow$ tan x = -1

= tan $\frac{-\pi }{4}$

$\Rightarrow x=n\pi$ - $\frac{\pi }{4}$ , when we put n = 0,1,2,3, ........ we get,

x = $\frac{-\pi }{4}$,$\frac{3\pi }{4}$,$\frac{7\pi }{4}$,$\frac{11\pi }{4}$,$\frac{15\pi }{4}$ ...... out of which $\frac{3\pi }{4}$ , $\frac{7\pi }{4}$ , belongs to $[0,2\pi ]$. Therefore two values