Question

plz answer and tell me! ​

Answer 1

Answer:

x = (2nπ ± 2 π / 3 )

Step-by-step explanation:

Given--->

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3 Cosx + 2 Sin² x = 0

To find--->

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Value of x

Solution--->

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3 Cos x + 2 Sin² x = 0

=> 2 Sin² x + 3 Cosx = 0

We have a formula

Sin² x + Cos² x = 1

Sin² x = 1 - Cos² x

Applying it here

=> 2 ( 1 - Cos² x ) + 3 Cos x = 0

=> 2 - 2 Cos² x + 3 Cos x = 0

=> - 2 Cos²x + 3 Cos x + 2 = 0

=> 2 Cos² x - 3 Cos x - 2 = 0

=> 2 Cos² x - ( 4 - 1 ) Cosx - 2 = 0

=> 2 Cos² x - ( 4 Cosx - Cos x ) - 2 = 0

=> 2 Cos² x - 4Cosx + Cosx - 2 = 0

=> 2Cosx ( Cosx - 2 ) +1(Cosx - 2 )= 0

=> ( Cosx - 2 ) ( 2 Cosx + 1 ) = 0

If Cosx - 2 = 0

Cos x = 2 > 1 ( impossible)

Because maximum value of Cosx is 1

If 2 Cosx + 1 = 0

=> 2 Cosx = - 1

=> Cosx = - 1 / 2

=> Cosx = Cos (2 π/3)

=> Cosx = Cos ( 2nπ ± 2 π / 3 )

=> x = 2nπ ± ( 2π / 3)