Question

17. Solve each of the following equations for 0 (a) 4 cos 2x + 2 sin x = 3 (b) sin 2x = sin2x (c) 7 sin x cos x = 2

Answer 1

a) A is a hard one but here goes...

$4cos2x + 2sinx = 3\\4cos2x + 2sinx -3 = 0\\4(1-sin^2x) +2 sinx - 3 = 0\\let w = sinx\\4(1-2w^2) + 2w -3 = 0\\8w^2 + 2w - 3 = 0\\factor\\\\(2w-1)(4w+3)=0\\\\w = sinx = \frac{1}{2},\frac{-3}{4\\}\\x= arcsin(\frac{1}{2}) = \frac{\pi}{6} + 2k\pi \\\\and \\x =arcsin(\frac{-3}{4} )+2k\pi$

b)

$sin2x = sin2x\\sin2x - sin2x = 0$

x = all real numbers

c)

$7sinx*cosx =2\\7 * \frac{sin2x}{2} = 2\\7*sin2x = 4\\sin2x = \frac{4}{7}\\2x = arcsin(4/7)\\x = \frac{arcsin\frac{4}{7} }{2}+2\pi k$ where k is an integer.