Hey guys! Does anybody know the answer to tan2x-2cosx=0

If the equation is tan(2x) - 2 cos(x) = 0:

We have

tan(2x) = sin(2x) / cos(2x)

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

So, rewriting the equation gives

tan(2x) - 2 cos(x) = 0

2 sin(x) cos(x) / cos(2x) - 2 cos(x) cos(2x) / cos(2x)= 0

2 cos(x) • (sin(x) - cos(2x)) / cos(2x) = 0

The left side is 0 whenever the numerator is 0:

2 cos(x) = 0   or   sin(x) - cos(2x) = 0

cos(x) = 0   or   sin(x) - (1 - 2 sin²(x)) = 0

cos(x) = 0   or   2 sin²(x) + sin(x) - 1 = 0

cos(x) = 0   or   (2 sin(x) - 1) (sin(x) + 1) = 0

cos(x) = 0   or   2 sin(x) - 1 = 0   or   sin(x) + 1 = 0

cos(x) = 0   or   sin(x) = 1/2   or   sin(x) = -1

Solving each case gives 3 families of solutions:

[x = π/2 + 2nπ   or   x = -π/2 + 2nπ]   or

[x = π/6 + 2nπ   or   x = 5π/6 + 2nπ]   or

[x = -π/2 + 2nπ]

where n is any integer. The last family is already accounted for in the first, so

x = π/2 + 2nπ   or   x = -π/2 + 2nπ   or   x = π/6 + 2nπ   or   x = 5π/6 + 2nπ

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

srry idk

step-by-step explanation: