If $0 \leq x \leq 2 π$ , then the number of solution of $3 (sin x + cos x) - 2 ({sin}^{2} x + {cos}^{2} x) = 8$ is:

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $0$
$3 (s i n x + c o s x) - 2 (s i n^{2} x + c o s^{2} x) = 8$
$3 (s i n x + c o s x) - 2 (1) = 8$
$3 (s i n x + c o s x) = 10$
$s i n x + c o s x = \frac{10}{3}$
Therefore,
$- 1 \leq s i n θ \leq 1$ and $- 1 \leq c o s x \leq 1$
Implies that,
$- 2 \leq s i n x + c o s x \leq 2$
Hence no value of $x$ exist.
Such that $s i n x + c o s x = \frac{10}{3}$
Therefore, it has no solution.

Suggest Corrections

Similar questions

0 \leq x \leq 2 π

3 (sin x + cos x) - 2 ({sin}^{3} x + {cos}^{3} x) = 8

The number of solution of the equation $s i n x + s i n 2 x + s i n 3 x = c o s x + c o s 2 x + c o s 3 x, 0 \leq x \leq 2 π$ is

0 \leq x < 2 π

x

cos x + cos 2 x + cos 3 x + cos 4 x = 0

2 s i n^{2} x + 3 s i n x - 2 > 0

x^{2} - x - 2 < 0

- 1 < x < 2

x^{2} - x - 2 < 0

- 1 < x < 2

\begin{matrix} List - I & List - II (I) & Number of solutions of the equation & (P) & 0 e^{x} + e^{- x} = tan x \forall x \in [0, \frac{π}{2}) (II) & Number of solutions of the equations & (Q) & 1 x + y = \frac{2 π}{3} and cos x + cos y = \frac{3}{2} is (III) & Number of solutions of the equation & (R) & 2 cos x + 2 sin x = 1, x \in [0, 2 π) is (IV) & Number of solutions of the equation & (S) & Infinite (\sqrt{3} sin x + cos x)^{\sqrt{\sqrt{3} sin 2 x - cos 2 x + 2}} = 4 is \end{matrix}

Join BYJU'S Learning Program