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) \( \frac{1}{\cos ^{3} x}+\tan ^{3}-x \)

Alright, I need to solve these trigonometric equations step by step. Let's start with the first set of equations: 1) If \( 2 \quad \cos +1=0 \) and \( \sin \alpha<0 \), determine the value of \( I \). First, let's interpret the equation \( 2 \quad \cos +1=0 \). It seems like there's a formatting issue here. It might be intended to be \( 2 \cos +1=0 \). So, I'll proceed with that assumption. So, \( 2 \cos \alpha +1=0 \). Let's solve for \( \cos \alpha \): \( 2 \cos \alpha = -1 \) \( \cos \alpha = -\frac{1}{2} \) Now, we're given that \( \sin \alpha < 0 \). Since \( \cos \alpha = -\frac{1}{2} \), and \( \sin \alpha \) is negative, this suggests that \( \alpha \) is in the third quadrant where both sine and cosine are negative. Now, let's find \( \tan \alpha \): \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \) But we don't have the value of \( \sin \alpha \) yet. To find \( \sin \alpha \), we can use the Pythagorean identity: \( \sin^2 \alpha + \cos^2 \alpha = 1 \) Plugging in \( \cos \alpha = -\frac{1}{2} \): \( \sin^2 \alpha + \left(-\frac{1}{2}\right)^2 = 1 \) \( \sin^2 \alpha + \frac{1}{4} = 1 \) \( \sin^2 \alpha = 1 - \frac{1}{4} \) \( \sin^2 \alpha = \frac{3}{4} \) Since \( \sin \alpha < 0 \), we take the negative square root: \( \sin \alpha = -\frac{\sqrt{3}}{2} \) Now, calculate \( \tan \alpha \): \( \tan \alpha = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \) So, \( \tan \alpha = \sqrt{3} \). Now, let's evaluate the expressions: (a) \( \tan \alpha = \sin \alpha \) We have \( \tan \alpha = \sqrt{3} \) and \( \sin \alpha = -\frac{\sqrt{3}}{2} \). Clearly, \( \sqrt{3} \neq -\frac{\sqrt{3}}{2} \), so this statement is **false**. (b) \( 2 \sin^2 \alpha - \cos^2 \alpha = \) We have \( \sin^2 \alpha = \frac{3}{4} \) and \( \cos^2 \alpha = \frac{1}{4} \). Plugging in: \( 2 \times \frac{3}{4} - \frac{1}{4} = \frac{6}{4} - \frac{1}{4} = \frac{5}{4} \) So, \( 2 \sin^2 \alpha - \cos^2 \alpha = \frac{5}{4} \). Now, moving on to the second set of equations: 2) \( \sqrt{2} \tan x +1=0 \) and \( \cos x > 0 \) Let's solve for \( \tan x \): \( \sqrt{2} \tan x = -1 \) \( \tan x = -\frac{1}{\sqrt{2}} \) Now, since \( \cos x > 0 \), and \( \tan x = \frac{\sin x}{\cos x} \), this implies that \( \sin x \) must be negative because \( \tan x \) is negative and \( \cos x \) is positive. So, \( \alpha \) is in the fourth quadrant where \( \cos x > 0 \) and \( \sin x < 0 \). Now, let's find \( \sin x \): Using the identity \( \sin^2 x + \cos^2 x = 1 \): \( \sin^2 x + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 \) \( \sin^2 x + \frac{1}{2} = 1 \) \( \sin^2 x = 1 - \frac{1}{2} = \frac{1}{2} \) Since \( \sin x < 0 \), \( \sin x = -\frac{1}{\sqrt{2}} \) Now, let's evaluate the expressions: (a) \( \sqrt{3} \cdot \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} \) Plugging in the values: \( \sqrt{3} \times \left(-\frac{1}{\sqrt{2}}\right) - \frac{\sqrt{2}}{\frac{1}{\sqrt{2}}} = -\frac{\sqrt{3}}{\sqrt{2}} - 2 = -\frac{\sqrt{6}}{2} - 2 \) So, \( \sqrt{3} \cdot \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} = -\frac{\sqrt{6}}{2} - 2 \) (b) \( \frac{1}{\cos^3 x} + \tan^3 x \) Plugging in the values: \( \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^3} + \left(-\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{\frac{1}{2\sqrt{2}}} + \left(-\frac{1}{2\sqrt{2}}\right) = 2\sqrt{2} - \frac{1}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{3}{2\sqrt{2}} \) So, \( \frac{1}{\cos^3 x} + \tan^3 x = \frac{3}{2\sqrt{2}} \)