# Solve the equation cos2x+sinx=0

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To solve the equation, we'll try to write it using just a single function.

We'll write cos 2x = cos (x+x) = cosx*cosx - sinx*sinx = (cos x)^2 - (sin x)^2

We'll substitute (cos x)^2 = 1 - (sin x)^2 in the formula of cos 2x:

cos 2x = 1 - (sin x)^2 - (sin x)^2

cos 2x = 1 - 2(sin x)^2

The equation will become:

1 - 2(sin x)^2 + sin x = 0

- 2(sin x)^2 + sin x + 1 = 0

We'll multiply by -1:

2(sin x)^2 - sin x - 1 = 0

We'll substitute sin x = t

2t^2 - t - 1 = 0

We'll apply the quadratic formula:

t1 = [1+sqrt(1+8)]/4

t1 = (1+3)/4

t1 = 1

t2 = -2/4

t2 = -1/2

So, sin x = t1

sin x = 1

x = arcsin (1) + k*pi

x = pi/2 + k*pi

sin x = -1/2

x = (-1)^(k+1) * arcsin (1/2) + k*pi

x = (-1)^(k+1) *(pi/6) + k*pi

k is an integer number.