How to solve this A-level pure math 1 question

i).Show that the equation 2 cos x = 3 tan x can be written as a quadratic equation in sin x
I found 2sin^2x+3sinx-2=0
Now the real problem
ii).
Solve the equation 2 cos 2y = 3 tan 2y, for 0
◦
≤ y ≤ 180
I used this cos(2x) = cos2(x) – sin2(x) = 1 – 2sin2(x) = 2cos2(x) – 1
and everything went so long and crazy. How it can be related to the first one. help

2cos(x) = 3tan(x)

2cos(x) = 3sin(x)/cos(x)

2cos²(x) = 3sin(x)

2cos²(x) - 3sin(x) = 0

2[1 - sin²(x)] - 3sin(x) = 0

2 - 2sin²(x) - 3sin(x) = 0

- 2sin²(x) - 3sin(x) + 2 = 0

2sin²(x) + 3sin(x) - 2 = 0 → you found this equation : ok



Conclusion:

Solving [2cos(x) = 3tan(x)] is similar solving [2sin²(x) + 3sin(x) - 2 = 0]

Adapt this result to your case. Suppose that: x = 2y

Solving [2cos(2y) = 3tan(2y)] is similar solving [2sin²(2y) + 3sin(2y) - 2 = 0]


Let's go:

2sin²(2y) + 3sin(2y) - 2 = 0 → let: sin(2y) = X → where - 1 ≤ X ≤ 1

2X² + 3X - 2 = 0

2X² + (4X - X) - 2 = 0

2X² + 4X - X - 2 = 0

(2X² + 4X) - (X + 2) = 0

2X(X + 2) - (X + 2) = 0

(X + 2)(2X - 1) = 0


First case: (X + 2) = 0 → X = - 2 → no possible because the condition

Second case: (2X - 1) = 0 → 2X = 1 → X = 1/2 → possible


sin(2y) = 1/2

2y = π/6 → y = π/12 → y = 15°

2y = π - (π/6) → 2y = 5π/6 → y = 5π/12 → y = 75°

→ Solution = {15° ; 75° }

If you sub x=2y in the equation in (i) you get the equation in (ii).

So solve 2sin^2x+3sinx-2=0 to give sinx=1/2 or sinx=-2 which in not
possible so sinx=1/2 giving x=30° or 150° or 390 etc
But x=2y so 2y=30 giving y=15° or 2y=150 giving y=75° and others are
outside the required range.

2cosx = 3tanx

=> 2cosx = 3sinx/cosx

=> 2cos²x = 3sinx

=> 2(1 - sin²x) = 3sinx

i.e. 2sin²x + 3sinx - 2 = 0.....correct.