sin^2è + 2sinè +1 =0
that e symbol is suppose to be theta
treat sine theta as x, and it get's really easy!
x^2+2x+1=0
Factor,
(x+1)(x+1)=0
now replace sine theta:
(sine theta + 1)^2=0
theta=sine^-1(0)
theta=0
Cheers,
Houdini
I again disagree with Houdini
He is ok to the end of
(sinØ + 1)^2 = 0
sinØ + 1 = 0
sinØ = -1
Ø = 270° or 3π/2 radians
Houdini's answer does not satisfy the equation.
Reiny's right -.-
tan pi/5=
To solve the equation sin^2è + 2sinè + 1 = 0, we can use a basic algebraic technique called factoring.
Step 1: Rewrite the equation in a standard quadratic form.
sin^2è + 2sinè + 1 = 0 can also be written as (sinè + 1)(sinè + 1) = 0.
Step 2: Apply the zero product property.
According to the zero product property, if a product of factors equals zero, then at least one of the factors must equal zero.
Thus, sinè + 1 = 0.
Step 3: Solve for the variable.
To solve sinè + 1 = 0, subtract 1 from both sides:
sinè = -1.
Step 4: Find the values of è.
To find the values of è, we can rely on the unit circle or the reference angles of sine.
On the unit circle, the sine function is -1 at two points: -π/2 and 3π/2, or in degrees, -90° and 270°.
Since the sin function is periodic with a period of 2π (or 360°), we can find all solutions by adding multiples of the period.
Thus, the solutions for sinè + 1 = 0 are:
è = -π/2 + 2πn, where n is an integer.
è = 3π/2 + 2πn, where n is an integer.
These are the values of è that satisfy the equation sin^2è + 2sinè + 1 = 0.