Question

Find the general solution of the equation: 3 sin⁴ x + cos⁴ x = 1.

Answer 1

Answer:

x=π/4+2πk,where k is any integer

or

x=0+2πk,where k is any integer

Step-by-step explanation:

To find the general solution of the equation: 3 sin⁴ x + cos⁴ x = 1 try to convert the given equation in terms of cos or sine

As we know that sin²x=1-cos²x

so sin⁴x=(1-cos²x)²

3 (1-cos²x)² + cos⁴ x = 1

3[1+cos⁴ x-2cos²x]+cos⁴ x = 1

3+3cos⁴ x-6cos²x+cos⁴ x = 1

4cos⁴ x -6cos²x+2=0

2cos⁴ x-3cos²x+1=0

2cos⁴ x-2cos²x-cos²x+1=0

2cos²x(cos²x-1)-1(cos²x-1)=0

(cos²x-1)(2cos²x-1)=0

(cos²x-1)=0

cos²x=1

cos x=1

x= cos⁻¹ (1)

x=0

or

2cos²x-1=0

2cos²x=1

cos²x=1/2

cos x=±1/√2

x = cos⁻¹(±1/√2)

x=π/4

so solution of the equation are x=0,π/4

So general solution of the equation are

x=π/4+2πk,where k is any integer

or

x=0+2πk

Answer 2

Solution :

3sin⁴x + cos⁴x = 1

=> 3( sin² x )² + ( cos² x )² = 1

=> 3[(1-cos2x )/2]²+[(1 +cos2x)/2]² = 1

=> 3[(1+cos²2x-2cos2x)/4]

+ [(1+cos²2x+2cos2x)/4] = 1

=> 3+3cos²2x-6cos2x

+1+cos²2x+2cos2x = 4

=> 4cos²2x - 4cos2x = 0

=> 4cos2x ( cos2x - 1 ) = 0

Therefore ,

4cos2x = 0 or cos2x - 1 = 0

=> cos 2x = 0 OR cos 2x = 1

1 - 2sin² x = 0 or 1 - 2sin²2x = 1

=> 2sin²x = 1 or 2sin² x = 0

=> sin² x = 1/2 or sin² x = 0

x = nπ ± ( π/4 ) or x = nπ , n € Z

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