Question

plz solve for θ :-

2sin²θ= 3cosθ ; where 0≤ θ ≤ 2π

{Class XI TRIGONOMETRY}​

Answer 1

Given Expression,

2sin²∅=3cos∅

»2sin²∅-3cos∅=0

We know that,

sin²x+cos²x=1→sin²x=1-cos²x

»2(1-cos²∅)-3cos∅=0

»2-2cos²∅-3cos∅=0

»2cos²∅+3cos∅-2=0

On comparing with ax²+bx+c=0,

a=2,b=3 and c= -2

Discriminant,D:

b²-4ac

= (3)²-4(2)(-2)

= 9+16

=25

Now,

$\sf{cos \: x = \frac{ - b \pm \sqrt{d} }{2a} } \\ \\ \: \: \: \: \: \: \: \: \: = \frac{ - 3 \pm \: \sqrt{25} }{4} \\ \\ \: \: \: \: \: \: \: \: = \frac{ - 3 \pm5}{4} \\ \\ \: \: \: \: \: \: \: = \frac{5 - 3}{4} \sf{or} \: \frac{ - 5 - 3}{4} \\ \\ \implies \boxed{ \sf{cos \: x = \frac{1}{2} \: or \: - 2 }}$

The principle solutions are 1/2 and -2

Implies,

cos x= π/3

General Solution:

x = nπ±y

»x= nπ±π/3

The range of the solutions is (-1,1)