# If 2sin^2θ + 5sinθ - 3 = 0. Find the value of tan^4θ + tan^2θ +9

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### 1 Answer

Let set `u = sintheta`

So we will have:

`2u^(2) +5u - 3 = 0`

Factor the left side.

`(2u - 1)(u + 3) = 0`

Equate each factor to zero.

`2u - 1 = 0` ;`u + 3 = 0`

Add 1 on both sides, to isolate the 2u on left side.

`2u = 1`

Divide both sides by 2 to isolate the u on left side.

` ` `u = 1/2`

Plug-in u = sinx.

`sintheta = 1/2`

Therefore, theta = pi/6, 5pi/6.

Note that for u + 3 = 0, we will have u = -3. Which will leads to

sintheta = -3. Which has no solution, since range of sinx is from -1 to 1.

Now, we plug-in the theta we got.

`tan^(4)(pi/6) + tan^(2)(pi/6) + 9 = (sqrt(3)/3)^4 + (sqrt(3)/3)^2 + 9`

`9/81 + 3/9 + 9 = 9/81 + 27/81 + 81/9 = 85/9`

Therefore, the value for theta = pi/6 is 13.

Plug-in theta = 5pi/6.

`tan^(4)((5pi)/6) + tan^(2)((5pi)/6) + 9 = (-sqrt(3)/3)^4 + (-sqrt(3)/3)^2 + 9 = 85/9`

Therefore **the value is 85/9**.