Math, asked by harshith8143, 1 month ago

The exhaustive set of values of a' for which the equation sin^4 x+cos^4 x-sin2x +a=0 has a solution is​

Answers

Answered by abhipsha02755
2

Answer:

Step-by-step explanation:

sin  

4

x+cos  

4

x+sin2x+α=0

⇒(sin  

2

x+cos  

2

x)  

2

−2sin  

2

xcos  

2

x+sin(2x)+α=0

⇒1−2sin  

2

xcos  

2

x+sin(2x)+α=0⇒1−  

2

sin  

2

(2x)

+sin(2x)+α=0

⇒2−sin  

2

(2x)+2sin(2x)+2α=0

⇒sin  

2

(2x)−2sin(2x)−(2α+2)=0

Let y=sin2x, so that y=[−1,1], \sin ce −1≤sin2x≤1

Thus, y  

2

−2y−(2α+2)=0

⇒y=1±  

3+2a

 

But, −1≤y≤1, so:

⇒−1≤1±  

3+2a

≤1

⇒−2≤±  

3+2a

≤0

\sin ce, the range is from -2 to 0,  then ignore the positive root

⇒−2≤−  

3+2a

≤0

⇒0≤3+2a≤4

⇒−3≤2a≤1

⇒−  

2

3

≤a≤  

2

1

 

Therefore, Answer is −  

2

3

≤a≤  

2

1

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