How do I solve the equation, (cos(x))^2 = 1/25, in the interval of [0,2π]?

Question

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Answer 1

How do I solve the equation, (cos(x))² = 1/25, in the interval of [0,2π]? Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Detailed Solution:

Given/Known: (cos(x))² = 1/25 Interval of [0,2π] x = ?? radian

(cos(x))² = 1/25 <== Take square root of both sides

cos(x) = ±1/5 ==> cos(x) = –1/5 (Quadrant 2 & 3) and cos(x) = 1/5 (Quadrant 1 & 4) ==> 4 Solutions

Four solutions in the Interval of [0,2π]

Quadrant 1: cos(x) = 1/5 ==> x = cos^-1 (1/5) = 1.369438406 radian

Quadrant 2: cos(x) = –1/5 ==> x = π – cos^-1 (1/5) = 1.772154248 radian

Quadrant 3: cos(x) = –1/5 ==> x = π + cos^-1 (1/5) = 4.51103106 radian

Quadrant 4: cos(x) = 1/5 ==> x = 2π – cos^-1 (1/5) = 4.913746901 radian

x = (Quadrant 1, Quadrant 2, Quadrant 3, Quadrant 4)

x = (cos^-1 (1/5), π – cos^-1 (1/5), π + cos^-1 (1/5), 2π – cos^-1 (1/5)) radian <== Exact Solution

x = (1.369438406, 1.772154248, 4.51103106, 4.913746901) radian <== Approximated Solution

Answer 2

Clearly cos (x) = 1/5 or cos (x) = - 1/5

x = cos^-1 (1/5) = 78.5 ° = 1.37 radians in the first quadrant

If 0 < α < π/2

then cos (α) = cos (2π-α)

So a second solution is 2π - 1.37

In the 2nd and 3rd quadrants cos is negative, so by symmetry there are two more solutions

π - 1.37

π + 1.37