# solve trigonometry using algebra2 sin 2x + 2+ 3cos x = -3sin x i don't have a clue!

*print*Print*list*Cite

We'll solve this equation algebraically.

We'll shift all terms to one side:

2sin2x + 3(sinx+cosx) + 2 = 0

We'll use double angle identity: sin 2x = 2sin x*cos x

2*2sin x*cos x + 3(sinx+cosx) + 2 = 0

We'll note sin x + cos x = y.

We'll raise to square and we'll get:

(sin x + cos x)^2 = y^2

(sin x)^2 + (cos x)^2 + 2sin x*cos x = y^2

But (sin x)^2 + (cos x)^2 = 1:

1 + 2sin x*cos x = y^2

2sin x*cos x = y^2 - 1

We'll re-write the given equation in y:

2*(y^2 - 1) + 3(y) + 2 = 0

We'll remove the brackets:

2y^2 - 2 + 3y + 2 = 0

We'll eliminate like terms:

2y^2 + 3y = 0

We'll factorize by y:

y(2y + 3) = 0

We'll put y = 0

But y = sin x + cos x:

sin x + cos x = 0

We'll divide by cos x:

tan x + 1 = 0

tan x = -1

x = -pi/4 + kpi

2y + 3 = 0

2y = -3

y = -3/2

But the range of values of y is [-2;2].

The maximum value of the sum is approached for x = pi/4: sin x + cos x = 1 + 1 = 2

The same with the minimum of the sum: sin x + cos x = -1-1 = -2

**The only solution of the equation is: ****x = -pi/4 + kpi.**