40 - 10cos(2x) + 40sin(2x)=0 How can I solve the following equation because I'm really stuck?

Question

40 - 10cos(2x) + 40sin(2x)=0

I need to solve for x and I really want to see and understand the steps needed to reach the correct answer.

Thanks in advance

Answer 1

I see A*cos(box)+B*sin(box)=k

then always it is the same as [√(A²+B²)]*cos( box-tan^-1(B/A))=k

here [√(40²+10²)]*cos(2x-tan^-1(40/-10))= -40

2x=[±cos^-1(-40/ [√(40²+10²)])]+tan^-1(40/-10)+2πk,,,,,,k any integer

impotant note on tan^-1(40/-10), principle value is --1.33 in fourth quadrant, but if y=40, x=-10

then angle is in second quadrant. must use tan^-1(40/-10)= -1.35+π=1.82 radians

x=2.36+πk is first set of answers

and x= -.54+πk is second set

when checking 2x=4.72+2πk or 2x=-1.08+2πk

Answer 2

Simplify and rewrite into

4(1 + sin(2x)) = cos(2x)

Square both sides

16(1 + 2sin(2x) + sin²(2x)) = cos²(2x)

Rewrite right hand side using "sin²(z) + cos²(z) = 1"

16(1 + 2sin(2x) + sin²(2x)) = 1 - sin²(2x)

Put into canonical form for quadratic equations

17sin²(2x) + 32sin(2x) + 15 = 0

Solve as quadratic equation

sin(2x) = (-32 ± √(32² - 4×17×15))/34

Gives two sets of solutions:

1)

sin(2x) = -1

2x = 3π/2 + 2kπ, for any integer k

x = 3π/4 + kπ, for any integer k

2)

sin(2x) = -15/17

2x = 5.2 + 2kπ, for any integer k

x = 2.6 + kπ, for any integer k