Vì `0^o < x < 90^o` nên `\sin x > 0`; `\cos x > 0`

`a)` `3\cos x + 2\sin(90^o - x) = 4,15` $(*)$

Vì `0^o < x < 90^o => \sin(90^o - x) = \cos x`, thay vào $(*)$ được:

`3\cos x + 2\cos x = 4,15`

`5\cos x = 4,15`

`\cos x = 0,83`

`x \approx 33,9^o` 

Vậy `x \approx 33,9^o` 

`b)`

`2\sin^2 x + \cos^2 x = 1,8281` $(*)$

Vì `\sin^2 x + \cos^2 x = 1 => \cos^2 x = 1 - \sin^2 x`, thay vào $(*)$ được:

`2\sin^2 x + (1 - \sin^2 x) = 1,8281`

`\sin^2 x + 1 = 1,8281`

`\sin^2 x = 0,8281`

`\sin x = \sqrt{0,8281} = 0,91` (do `\sin x > 0`)

`x \approx 65,51^o`

 Vậy `x \approx 65,51^o`

`c)` `\cos^2 x - \sin^2 x = 0,5` $(*)$

Vì `\sin^2 x + \cos^2 x = 1 => \sin^2 x = 1 - \cos^2 x`, thay vào $(*)$ được:

`\cos^2 x - (1 - \cos^2 x) = 0,5`

`2\cos^2 x - 1 = 0,5`

`2\cos^2 x = 1,5`

`\cos^2 x = 0,75 = (3)/(4)`

`\cos x = (\sqrt{3})/(2)` (do `\cos x > 0`)

`x = 30^o`

Vậy `x = 30^o`

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