Find all x for this equation 2cosx-3=-5, 0<_x<_2π

Question

The range is 0 is less than equal x and x is less than equal 2π.

Solve this according to the rules of unit circle.

Answer 1

2cosx -3 = -5 add 3 to both sides

2cosx = -5 +3 = -2

divide by 2

cosx =-2/2 =- 1

cosx =-1

x = pi radians or 180 degrees

x = cos^-1(-1)

one way to find x is use a calculator with an inverse trig function. Punch in -1 then punch the inverse cosine key

or sketch a unit circle and realize that cosine = adjacent side over hypotenuse = -1

cosines are negative only in the 2nd and 3rd quadrants. -1 is the minimum possible value of cosine of any angle. that's the boundary of quadrant 2 and 3, where x=pi or 180

Or draw the cosine curve, It equals 1 when x = 0, and reaches a minimum -1 when x= 180 or pi, half the period of 2pi

Answer 2

For this kind of problem, we can first solve for cos(x), and then reference the unit circle to demine which values of x correspond to the equation.

We start with.

2cos(x) - 3 = -5

2cos(x) = -2

cos(x) = -1

So, we just need to find the angles on the unit circle where cos(x) = -1.

Remember that cos is the x-coordinate on the unit circle.

Therefore, x = π

This is the only value of x since x is bounded between 0 and 2π. (one revolution around the unit circle)