# Determine x if sin2x=-(sinx)^2.

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### 3 Answers

sin2x = -(sinx)^2

We know that:

sin 2x = 2sinx *cosx

==> 2sinx cosx = -(sinx)^2

==> (sinx)^2 + 2sinx cos x = -

Factor sinx :

==> sinx (sinx + 2cosx) = 0

==> sinx = 0 ==> x1 = n*pi (n=0,1,..)

=> sinx + 2cos x = 0

Let us divide by cosx

==> tanx + 2 = 0

==> tanx = -2 ==> x2 = arctan(-2)= -arctan(2)

The first step is to re-write the expression, using the formula for the double angle:

sin 2x = 2sinx*cosx.

We'll re-write now the entire expression, moving all terms to one side.

(sin x)^2+2sin x * cos x = 0

We'll factorize and we'll get:

sin x * (sin x + 2cos x) = 0

We'll put each factor from the product as 0.

sin x = 0

We notice that it is an elementary equation.

x = (-1)^k*arcsin 0 + k*pi

x = k*pi

We'll put the next factor as zero.

sin x + 2cos x = 0

This is a homogeneous equation, in sin x and cos x.

We'll divide the entire equation, by cos x.

sin x / cos x + 2 = 0

But the ratio sin x / cos x = tg x.

We'll substitute the ratiosin x / cos x by tg x.

tg x + 2 = 0

x = arctg(-2 ) +k*pi

x = pi - arctg2 + k*pi

x = pi*(k+1) - arctg 2

The x values for the expression to be true are:

**{k*pi}U{pi*(k+1) - arctg 2}**

To solve for x if sin2x = - (sinx)^2.

Rewrite the equation as:

sin2x +(sinx)^2 = 0.

2sixcosx +sinx = 0. Factorise sinx.

sinx(2cosx +1) = 0

sinx =0 , 2cosx = -1

For sinx =0, x =npi , n = 0, 1,2...

For 2cosx = -1, cosx = -1/2 . x = 2npi + pi/3 or x = 2npi -pi/3 , n=0, 1,2 .....