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Trigonometric Equations

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3sec2x+4cos2x=7

 

please could someone give me a clear step by step way of how to solve??

asked Feb 13, 2013 in TRIGONOMETRY by chrisgirl Apprentice

2 Answers

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3sec2x + 4cos2x = 7

Reciprocal identities: secθ = 1/cosθ

3(1/cos2x) + 4cos2x = 7

Rewrite the expression with common denominator.

[3 + 4cos4x]/cos2x = 7

Multiply each side by cos2x.

[3 + 4cos4x] = 7cos2x

Subtract 7cos2x from each side.

4cos4x - 7cos2x + 3 = 0

Let cos2x = t

4t2 - 7t + 3 = 0

Now solve the equation using factor method.

4t2 - 4t - 3t + 3 = 0

4t(t - 1) -3(t - 1) = 0

Take out common factors.

(4t - 3)(t - 1) = 0

4t - 3 = 0 or t - 1 = 0

4t - 3 = 0 then t = 3/4

t - 1 = 0 then t = 1

But t = cos2x

So, cos2x = 1 or cos2x = 3/4

cos2x = 1⇒cos(x) = √1 = ±1

If cosx = 1 ⇒ cosx = cos0 = cos360 then x = 0 or 360 = 2π

If cosx = -1⇒ cosx = cos180 then x = 180 = π.

Take cos2x = 3/4 ⇒cos(x) = √(3/4) = ±√3/2 = ±0.866

If cosx = 0.866⇒cosx = cos390 then x = 390 = π/6 + 2π*n    [n is an integer]

If cosx = -0.866 ⇒cosx = cos210 then x = 210 = π/6 + π*n   [n is an integer]

General solutions x = π/6 + 2π * n, 5π/6 + 2π * n, 7π/6 + 2π * n, 11π/6 + 2π * n.

Therefore interval [0, 2π], x = 0, pi, pi/6, 5pi/6, 7pi/6 , 11pi/6.

answered Feb 13, 2013 by britally Apprentice

The solutions are : x = 2nπ,

                         x = π(2n - 1) or x = π(2n + 1),

                         x = π(2n - 5/6) or x = π(2n + 5/6), and

                         x = π(2n - 1/6) or x = π(2n + 1/6), where n is an integer.

0 votes

The trigonometric equation is 3sec2x + 4cos2x = 7.

Using reciprocal identities: secθ = 1/cosθ.

3(1/cos2x) + 4cos2x = 7

3 + 4cos4x = 7cos2x.

4cos4x - 7cos2x + 3 = 0

Let, cos2x = t.

4t2 - 7t + 3 = 0

By factor by grouping.

4t2 - 4t - 3t + 3 = 0

4t(t - 1) -3(t - 1) = 0

(4t - 3)(t - 1) = 0

Apply zero product property.

4t - 3 = 0 or t - 1 = 0

4t = 3 or t = 1

t = 3/4 or t = 1.

Put, t = cos2x.

cos2x = 3/4 or cos2x = 1

cos x = ± √3/2 or cos x = ± 1.

  • cos x = - √3/2.

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

The function cos(x) is negative in second and third quadrant.

cos x = cos 5π/6

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer. 

x = 2nπ ± 5π/6

x = π(2n - 5/6) or x = π(2n + 5/6).

  • cos x = √3/2.

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

The function cos(x) is positive in first and fourth quadrant.

cos x = cos π/6

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer. 

x = 2nπ ± π/6

x = π(2n - 1/6) or x = π(2n + 1/6).

  • cos x = - 1.

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

The function cos(x) is negative in second and third quadrant.

cos x = cos π

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer. 

x = 2nπ ± π

x = π(2n - 1) or x = π(2n + 1).

  • cos x = 1.

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

The function cos(x) is positive in first and fourth quadrant.

cos x = cos 0

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

x = 2nπ ± 0

x = 2nπ.

The solutions are : x = 2nπ,

                         x = π(2n - 1) or x = π(2n + 1),

                         x = π(2n - 5/6) or x = π(2n + 5/6), and

                         x = π(2n - 1/6) or x = π(2n + 1/6), where n is an integer.

answered Jul 8, 2014 by lilly Expert

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