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

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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}.**

cost+1 = 2sint +sin2t. We proceed to solve for t.

We know sin2t = 2sint cost.

Therefore , the given equation becomes:

cost+1 = 2sint +2sint*cost.

cost+1 = 2sint(1+cost).

Subtract cost+1 from both sides:

0 = 2sint(1+cost) - (1+cost).

0 = 2sint(1+cost) - 1(1+cost).

0 = (cost+1)(2sint-1).

We equate the factors to zero.

cost+1 = 0. Or 2sint -1 = 0.

cost = -1, t = pi or t = (2n+1)pi

2sint = 1, sint = 1/2. t = npi+(-1)^n*pi/6.