Use inverse functions

Question

2 cos^2 x +7 sin x =5

Answer 1

2 cos² x + 7 sin x = 5

 

-2(1 - sin² x) - 7 sin x + 5 = 0

 

2 sin² x - 7 sin x + 3 = 0

 

(2 sin x - 1)(sin x - 3) = 0

 

sin x = 1/2. (The alternative, sin x = 3 has no solution as sin x has range -1 ≤ sin x ≤ 1)

 

In the interval 0 ≤ x < 2π, there are two solutions, x = π/6 or x = 5π/6.

 

The full solution set is thus

 

x = 2kπ + π/6 or x = 2kπ + 5π/6 ; k ∈ Z.

Answer 2

This program can be converted to an ordinary quadratic one involving only sin functions by using the identity sin^2 + cos^2 = 1.   So  2cos2^2 = 2 (1-sin^2x).

 

2(1-sin^2 x) + 7 sin x -5 = 0;

 

-2sin^2 x + 7sin x -3 = 0

2sin²x - 7 sin x +3 = 0

 

factoring yields 

(2sin x - 1)(sinx - 3) = 0;

 

sin x -3 = 0 has on solutions because sin x is always between -1 and 1.

2sin x -1 = 0 yields   sin x = 1/2.   The solutions are x = pi/6 plus multiples of 2pi  and x = 5pi/6 plus multiples of 2pi.

Checking into original equation we find that all of those solutions work.