What is t for cost+1=2sint+sin2t ?

To determine the angle t, we'll have to solve the
equation. We notice that one of the terms of the identity is the function sine of a
double angle.

We'll apply the formula for the double
angle:

sin 2a = sin (a+a)=sina*cosa +
sina*cosa=2sina*cosa

We'll replace 2a by 2t and we'll
get:

sin 2t = 2sin t*cos
t

We'll re-write the equation, moving all terms to one
side:

2sin t*cos t + 2sint - cost - 1 =
0

We'll factorize by 2sin t the first 2
terms:

2sint(cos t + 1) - (cos t + 1) =
0

We'll factorize by (cos t +
1):

(cos t + 1)(2sin t - 1) =
0

We'll set each factor as
zero:

cos t + 1 = 0

We'll add
-1 both sides:

cos t = -1

t =
arccos (-1)

t = pi

2sin t - 1
= 0

We'll add 1 both
sides:

2sin t = 1

sin t =
1/2

t = arcsin (1/2)

t =
pi/6

t =
5pi/6

The angle t has the following values:
{pi/6 ; 5pi/6 ; pi}.