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1) If \( 2 \quad \cos +1=0 \) and \( \sin \alpha<0 \) determis Winceit o ecilcutato with un and of a diaroum the value of I (a) \( \tan \alpha=\sin \alpha \) b) \( 2 \sin ^{2}-\cos ^{2}= \) (2) \( \sqrt{2} \tan x+1=0 \) and \( \cos 0 x>0 \) a) \( \sqrt{3} \cdot \sin \alpha-\frac{\sqrt{2}}{\cos \alpha} \) a) \( \frac{1}{\cos ^{3} x}+\tan ^{3}-x \)

Ask by Maxwell Li. in South Africa
Mar 11,2025

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1) For the first set of equations: (a) \( \tan \alpha = \sin \alpha \) is **false**. (b) \( 2 \sin^2 \alpha - \cos^2 \alpha = \frac{5}{4} \). 2) For the second set of equations: (a) \( \sqrt{3} \cdot \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} = -\frac{\sqrt{6}}{2} - 2 \). (b) \( \frac{1}{\cos^3 x} + \tan^3 x = \frac{3}{2\sqrt{2}} \).

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Bonus Knowledge

Let’s put on our math hats! First, to tackle your equations, it's helpful to know some key identities from trigonometry. For instance, the relationship between sine and cosine, like \( \sin^2 \alpha + \cos^2 \alpha = 1 \), can be a lifesaver. This means that while working through these equations, even if you get stuck, you can always convert between sine and cosine and still find a solution! Now, about those expressions you’re dealing with – noticing the signs and quadrants of each function is essential! For example, if you know \( \sin \alpha < 0 \), you're limited to the third and fourth quadrants, which helps you determine possible values for \( \alpha \). And remember, keeping track of the signs for secant, tangent, and cos as you solve can prevent squaring errors that lead to incorrect answers! Happy solving!

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