The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has:
A
infinite number of real roots
CHECK ANSWER
Explanation Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,
we get $${t^2} - 4t - 1 = 0$$
$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$
$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$
$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$
$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$
$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)
$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and
$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$
$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$
and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected
Hence given equation has no solution.
$$\therefore$$ The equation has no real roots.