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Find all solutions of the equation in the interval?

+1 vote
1)6 sin^2 x + 9 sin x + 3 = 0

2)7sinx= cot(−x)

3)4secx + 4tanx = 4
asked Jan 30, 2013 in TRIGONOMETRY by angel12 Scholar

6 Answers

+1 vote

6 sin2 x + 9 sin x + 3 = 0

Let sin2 x = y

6y2 + 9y + 3 = 0

Divide each side by 3.

2y2 + 3y + 1 = 0

Now solve the factor method.

2y2 + 2y + y + 1 = 0

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

Take out common factors.

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

2y +1 = 0 or y +1 = 0

2y +1 = 0

Subtract 1 from each side.

2y = -1

Divide each side by 2.

y = -1/2

And y + 1 = 0

Subtract 1 from each side.

y = -1

There fore y = -1/2 and y = -1

But y = sin x

So, sin x = -1/2 and sin x = -1

Trigonometric table in sin(11π/6) = -1/2 and sin(3π/2) = -1

There fore sin x = sin(11π/6) and sin x = sin(3π/2)

Cancel common terms.

x = 11π/6 and x = 3π/2

The equation in the interval is [3π/2, 11π/6]

answered Jan 30, 2013 by richardson Scholar
Let sin x = y

sin(x) = - 1/2 or sin(x) = - 1.

x = sin- 1(- 1/2) or x = sin- 1(- 1).

The function sin(x) has a period of , first find all solutions in the interval (0, 2π).

The function sin(x) is negative in third and fourth quadrant.

  • sin(x) = - 1/2.

In third Quadrant, πx ≤ 3π/2.

- 1/2 = - sin (π/6) = sin (π + π/6) = sin (7π/6).

In fourth Quadrant, 3π/2x ≤ 2π.

- 1/2 = - sin (π/6) = sin (2π - π/6) = sin (11π/6).

So, the general solutions are x = 2nπ + 7π/6, x = 2nπ + 11π/6 where n ∈ Z.

  • sin(x) = - 1.

In third Quadrant, πx ≤ 3π/2.

- 1 = - sin (π/2) = sin (π + π/2) = sin (3π/2).

In fourth Quadrant, 3π/2x ≤ 2π.

- 1 = - sin (π/2) = sin (2π - π/2) = sin (3π/2).

So, the general solution is x = 2nπ + 3π/2, where n ∈ Z.

The general solutions of 6 sin2 x + 9 sin x + 3 = 0 are x = 2nπ + 7π/6, x = 2nπ + 11π/6 and x = 2nπ + 3π/2, where n ∈ Z.

+1 vote

7sinx= cot(−x)

Odd/Even Identities: cot (–x) = –cot x

7sinx= - cot(x)

Ratio Identities: cot(x) = cos(x) / sin(x)

7sinx= - cos(x) / sin(x)

Multiply each side by sin(x).

7sin2x = - cos(x)

Add cos(x) to each side.

7sin2x + cos(x) = 0

Pythagorean Identities: sin2 θ + cos2 θ = 1⇒ sin2x = 1 - cos2x

7(1 - cos2x) + cos(x) = 0

7 - 7cos2x + cos(x) = 0

Multiply each side by negative one.

7cos2x - cos(x) - 7 = 0

Let cos x = y

7y2 - y - 7 = 0

Now solve the factor method.

Compare equation with standard from ax2+bx+c=0 and write the coefficients.

a = 7, b= -1 and c = -7.

The quadratic formula: x = [-b + √(b2 - 4ac)] / 2a

y = [-(-1)+ √((-1)2 - 4(7)(-7))] / 2(7)]

y = [1 + √(1 + 196)] / 14]

y = [1 + √(197)] / 14]

y = [1 + √(197)] / 14] or y = [1 - √(197)] / 14]

Simplify

y = 1.0739 (the nearest value is 1.0739=1) or y = 0.9311

But y = cos x

cos(x) = 1 or cos(x) = 0.9311

Trigonometric table of values cos 0= 0 and cos21 = 0.9311

cos x = cos 0 ⇒x = 0

cos x = cos21 ⇒x = 21

The equation in the interval is [0, 21]

answered Jan 30, 2013 by richardson Scholar
cos 90 = 0
+1 vote

4secx + 4tanx = 4

Take out common term 4.

4(secx + tanx) = 4

Divide each side by 4.

secx + tanx = 1

Reciprocal Identities: secθ = 1/cosθ and Ratio Identities: tanθ = sinθ/cosθ

1/cos(x) + sin(x)/cos(x) = 1

Remrite the expression with common denominator

(1 + sin(x)) / cos(x) = 1

Multiply each side by cos(x).

1 + sin(x) = cos(x)

.Subtract 'sin(x)' from each side.

1 = cos(x) - sin(x)

Apply square each side.

12 = [cos(x) - sin(x)]2

cos2x + sin2x - 2cos(x)sin(x )= 1

Pythagorean Identities: sin2 θ + cos2 θ = 1

1 - 2sin(x)cos(x) = 1

.Subtract 1 from each side.

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

Divide each side by -2.

sin(x)cos(x) = 0

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

Trigonometric table of values sin0 = 0 and cos 90= 0

sin(x) = sin(0) ⇒ x = 0

cos(x) = cos(90) ⇒ x = 90

The equation in the interval is [0, 90]

answered Jan 30, 2013 by richardson Scholar
good work!!!
0 votes
  • 2).

The trigonometric equation is image.

image

image

image

image

image

image

Let, image, then image.

image

image

image

image

image

image.

answered Jun 13, 2014 by lilly Expert
0 votes

Contd...........

Let, image,

image.

image, is a quadratic, use quadratic equation to find roots of the related quadratic function.

The solution of the quadratic equation : image is image.

Compare the equation image with image.

image.

image

image

image

image

image

Put  image.

image

image

put  image.

image.

image

image.

The general solution is image, where, image.

answered Jun 13, 2014 by lilly Expert
0 votes

(3).

Solve for x:

4 sec(x)+4 tan(x) = 4

Move everything to the left hand side.

Subtract 4 from both sides:

-4+4 sec(x)+4 tan(x) = 0

Factor the left hand side.

Factor constant terms from the left hand side:

4 (-1+sec(x)+tan(x)) = 0

Divide both sides by a constant to simplify the equation.

Divide both sides by 4:

-1+sec(x)+tan(x) = 0

Transform -1+sec(x)+tan(x) = 0 into a rational equation via the Weierstrass substitution.

Substitute y = tan(x/2). Then sin(x) = (2 y)/(y^2+1) and cos(x) = (1-y^2)/(y^2+1):

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

Divide both sides by a constant to simplify the equation.

Divide both sides by -2:

y/(y-1) = 0

Multiply both sides by a polynomial to clear fractions.

Multiply both sides by y-1:

y = 0

Perform back substitution on y = 0.

Substitute back for y = tan(x/2):

tan(x/2) = 0

Eliminate the tangent from the left hand side.

Take the inverse tangent of both sides:

x/2 = pi n  for  n element Z

Solve for x.

Multiply both sides by 2:

Answer: x = 2 pi n  for  n element Z

Source : http://www.wolframalpha.com/input/?i=solve+4secx+%2B+4tanx+%3D+4+for+x

answered Jun 13, 2014 by casacop Expert
edited Jun 13, 2014 by casacop

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