How to solve trigonometric functions such as

sin 2x sin x + cos x= 0

Use the double angle identity:

sin(2x) = 2 sin(x) cos(x)

Now rewrite

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

as

2 sin²(x) cos(x) + cos(x) = 0

Factor out cos(x) :

cos(x) (2 sin²(x) + 1) = 0

Consider the two cases,

cos(x) = 0   OR   2 sin²(x) + 1 = 0

Solve for cos(x) and sin²(x) :

cos(x) = 0   OR   sin²(x) = -1/2

Squaring a real number always gives a non-negative number, so the second case doesn't offer any real solutions. We're left with

cos(x) = 0

Cosine is zero for odd multiples of π/2, so we have

x = (2n + 1) π/2

where n is any integer.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

nada

step-by-step explanation: