Find all solutions in the interval 0 2π

Annie S.

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Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0
Math Analysis - Trigonometric Equation - Horizontal Shift - Sine
Click hre to learn how to find all of the solutions to a trigonometric equation over a specified interval, [0,2pi). We then verify the solutions using the GRAPH, WINDOW and TRACE features of the TI-84C. Horizontal Shift - Sine Function
Click hre to learn how to find all of the solutions to a trigonometric equation over a specified interval, [0,2pi). We then verify the solutions using the GRAPH, WINDOW and TRACE features of the TI-84C. Horizontal Shift - Sine Function

asked • 05/05/20

3 Answers By Expert Tutors

Ethan S. answered • 05/05/20

Let's throw the cosine function cos(x) into a variable - we'll say y = cos(x). Our expression now becomes

4y2 + 2y - 2 = 0

Factoring out the greatest common factor, 2, of the three terms yields the expression

2(2y2 + y - 1) = 0

To completely simplify, we can factor the expression 2y2 + y - 1 by grouping:

2y2 + y - 1

= 2y2 + 2y - y - 1

= 2y(y+1) - (y + 1)

= (2y - 1)(y + 1)

Our equation with the left hand side full factored now looks like

2(2y-1)(y+1) = 0

And we can solve for the two cases 2y - 1 = 0 and y + 1 = 0:

2y - 1 = 0 → 2y = 1 → y = 1/2

y + 1 = 0 → y = -1

Finally, we can substitute cos(x) for y, giving us the solutions cos(x) = 1/2 and cos(x) = -1. On the interval [0, 2π), cos(x) = 1/2 when x = π/3 and x = 5π/3, and cos(x) = -1 when x = π, so our solutions are x = π/3, x = 5π/3 and x = π

Factor this. GCF=2

2(2cos2x+cos x-1)=0

2(2cos x -1)(cos x+1)=0 Set each factor equal to zero and solve

2cos x -1=0 Add 1 to both sides

2cos x =1 Divide by 2

cos x =1/2

x=π/3

x=5π/3

cos x+1=0 Subtract 1 from both sides

cos x=-1

x=π

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Isolate the angle 2x, by following the reverse "order of operations".

Step 1: Add 1 to both sides:
#2cos^2(2x)=1#

Step 2: Divide both sides by 2:
#cos^2(2x) = 1/2#

Step 3: Take the square root of both sides:
#cos(2x) =(sqrt(2))/2 or cos(2x) =(-sqrt(2))/2#
(don't forget the positive and negative solutions!)

Step 4: Use inverse of cosine to find the angles:
#2x = cos^-1(sqrt(2)/2) or2x = cos^-1(-sqrt(2)/2) #

Step 5: Find angles that work:
#2x = pi/4 or 2x = (7pi)/4 or 2x=(3pi)/4 or 2x = (5pi)/4#

Step 6: Solve for x:
#x = pi/8, (7pi)/8, (3pi)/8, (5pi)/8# or .785, 5.5, 2.36, 3.93
(decimal approximations are seen on the graph below)

Explanation:

Use the Property: #sin^2x+cos^2x=1#

#2(1-cos^2x)+cosx-1=0#

#2-2cos^2x+cosx-1=0#

#2cos^2x-cosx-1=0#

#(2cosx+1)(cosx-1)=0#

#2cosx+1=0 or cosx-1=0#

#cosx=-1/2 or cosx=1#

#x=cos^-1(-1/2) or x=cos^-1 1#

#x=+- (2pi)/3 +2pin or x=0+2pin#

#n=0, x=(2pi)/3, -(2pi)/3,0#

#n=1, x=(8pi)/3, (4pi)/3,2pi#

#S={(2pi)/3,(4pi)/3}#

Solution:

Unlike normal solutions of algebraic equations with the number of solutions based on the degree of the variable, in trigonometric equations, the solutions are of two types, based on the different value of angle for the

trigonometric function, for the same solution.

Given (sin x)(cos x) = 0

If (sin x)(cos x) = 0 then either (sin x) = 0 or (cos x) = 0

For sin x = 0; x = 0, π , 2π.

For cosx = 0; x = π/2, 3π/2

Therefore, the solutions are 0,π, 2π π/2, 3π/2.