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General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2

Find the numb...

Question

Find the number of the solutions of the equation $1 + cos x + cos 2 x + sin x + sin 2 x + sin 3 x = 0$ which satisfy the condition $\frac{π}{2} < ∣ ∣ ∣ 3 x - \frac{π}{2} ∣ ∣ ∣ \leq π$ .

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Solution

The correct option is A $1$
$(1 + c o s (x) + c o s (2 x)) + (s i n x + s i n 2 x + s i n 3 x) = 0$
$(1 + c o s (x) + 2 c o s^{2} x - 1) + (s i n x + s i n 3 x + s i n 2 x) = 0$
$(2 c o s^{2} (x) + c o s (x)) + (2 s i n 2 x . c o s x + s i n 2 x) + 0$
$c o s (x) (2 c o s (x) + 1) + s i n 2 x (2 c o s (x) + 1) = 0$
$(c o s (x) + s i n 2 x) (2 c o s (x) + 1) = 0$
$2 c o s (x) + 1 = 0$
$c o s (x) = \frac{- 1}{2}$
Hence $x = \frac{2 π}{3}, \frac{4 π}{3}$
And
$c o s x + s i n 2 x = 0$
$c o s x + 2 s i n (x) c o s (x) = 0$
$c o s (x) [1 + 2 s i n (x)] = 0$
$c o s (x) = 0$ and $s i n (x) = \frac{- 1}{2}$
$x = \frac{π}{2}, \frac{3 π}{2}$ and $x = \frac{7 π}{6}, \frac{- π}{6}$ .
Now
$\frac{π}{2} < | 3 x - \frac{π}{2} | \leq π$
$| 3 x - \frac{π}{2} | > \frac{π}{2}$
Or
$3 x - \frac{π}{2} > \frac{π}{2}$ and $3 x - \frac{π}{2} < \frac{- π}{2}$
$x < 0$ and $x > \frac{π}{3}$ ...(i)
and
$| 3 x - \frac{π}{2} | \leq π$
$- π \leq 3 x - \frac{π}{2} \leq π$
$- 2 π \leq 6 x - π \leq π$
$\frac{- π}{6} \leq x \leq \frac{π}{3}$ ... (ii)
Hence from i and ii
$\frac{- π}{6} \leq x < 0$
Hence in the above range we get only one solution of the above equation, that is $= \frac{- π}{6}$ .
Hence answer is 1.

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Similar questions

The solution of the equation ${[s i n x + c o s x]}^{1 + s i n 2 x} = 2, - π \leq x \leq π$ is

{cos}^{2} x + cos x = {sin}^{2} x

x \in [0, π]

\frac{\sqrt{1 - {cos}^{2} x}}{sin x} + \frac{\sqrt{1 - {sin}^{2} x}}{cos x}

0 < x < \frac{π}{2}

\frac{sin x + sin 2 x}{1 + cos x + cos 2 x^{'}}

x \in (- \frac{π}{2}, \frac{π}{2}),

[sin x + cos x]^{1 + sin 2 x} = 2, - π \leq x \leq π

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