Question

the following equation has how many solution
$(i)\sqrt{3} \sin(x) + \cos(x) = 4 \\ ii) \sin x \: cosx = \frac{2}{3}$

Answer 1

Answer:

(i) 2

(ii) 4

Step-by-step explanation:

$\sqrt{3}sin(x) + cos(x) =4\\\sqrt{3}sinx = 4 -cos(x)\\ 3sin^{2}x = (4-cos(x))^{2}\\3(1-cos^{2}x) = 16 -8cos(x) + cos^{2}x\\ 4cos^{2}x - 8cos(x) +13 = 0$

$cos(x)=(8$±$\sqrt{64-208})/8$

$cos(x)=1$±$\sqrt{-144})/8$

$cos(x)=1$±$12i/8$

i.e  $1+3i/2$    $or$     $1-3i/2$     where $i^{2} = -1$

therefore $\sqrt{3} sin(x) + cos(x) = 4$  has two imaginary solutions