Solve \(2\sin^2 x + \cos x - 2 = 0\) for \(0 \leq x \leq 360^\circ\).

2(Sinx)^2 + Cosx - 2 = 0.
Recall (Sinx)^2 + (Cosx)^2 = 1. Therefore (Sinx)^2 = 1 - (Cosx)^2 Substitute this into the question above.
2(Sinx)^2 + Cosx - 2 = 0. 2(1 - (Cosx)^2) + Cosx - 2 = 0 Expand 2 - 2(Cosx)^2 + Cosx - 2 = 0 2 - 2 - 2(Cosx)^2 + Cosx = 0

2(Cosx)^2 + Cosx = 0 Multiply both sides by -1. 2(Cosx)^2 - Cosx = 0

Let p = Cosx 2p^2 - p = 0 Factorise p(2p - 1) = 0. Therefore p=0 or (p-1) = 0 p=0 or (p-1) = 0 p=0 or p = 0 +1. p=0 or p = 1 Recall p = Cosx
Therefore Cosx = 0 or 1. For 0 < x < 360
Cosx = 0, x = Cos inverse (0) , x = 90, 270 Cosx = 1, x = Cos inverse (1) , x = 0, 360
Therefore x = 0,90, 270 & 360 degrees.
Cheers.

Answered by olemakpadu | 2024-06-10

The solutions to the equation 2 sin 2 x + cos x − 2 = 0 for 0 ≤ x ≤ 36 0 ∘ are 6 0 ∘ , 9 0 ∘ , 27 0 ∘ , and 30 0 ∘ .
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Answered by olemakpadu | 2025-06-01