Question

please answer itttttt​

Answer 1

Theta is written as B.

Answer:

B = 60°.

Step-by-step explanation:

From the properties of trigonometric ratios :

• cos^2 A = 1 - sin^2 A

Here,

= > 3 sinB = 2 cos^2 B

= > 3 sinB = 2( 1 - sin^2 B )

= > 3 sinB = 2 - 2 sin^2 B

= > 3a = 2 - 2a^2 { to decrease length, let sinB = a }

= > 2a^2 + 3a - 2 = 0

= > 2a^2 + ( 4 - 1 )a - 2 = 0

= > 2a^2 + 4a - a - 2 = 0

= > 2a( a + 2 ) - ( a + 2 ) = 0

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

= > a + 2 = 0 Or 2a - 1 = 0

= > a = - 2 Or a = 1 / 2

= > sinB = - 2 Or sinB = 1 / 2

= > sinB ≠ - 2 Or sinB = sin30°

B = 30°.

Answer 2

Question :-- Solve 3sinA = 2cos²A .. Find A ?

Formula used :-

• cos²A= (1-sin²A)

Solution :--

→ 3sinA = 2cos²A

Putting value of cos²A in RHS,

→ 3sinA = 2 ( 1 - sin²A )

→ 3sinA = 2 - 2sin²A

→ 2sin²A + 3sinA - 2 = 0

Splitting the Middle Term now,

→ 2sin²A + 4sinA - sin A - 2 = 0

→ 2sinA(sinA + 2) - 1(sinA + 2) = 0

→ (2sin A - 1)(sin A + 2) = 0

Putting both Equal to 0 now,

→ 2sin A - 1 = 0. or, sin A + 2 = 0

→ Sin A = 1/2. SinA = (-2)

As , (-1) ≤ SinA ≤ 1 .

→ SinA = (-2) = Not Possible.

So,

→ SinA = 1/2

→ A = ( π/6 or 5π/6 ) if the interval is [0,2π)

or

→ A = (π/6 + 2πk or , 5π/6 + 2πk)..