1) If $$ 2 \quad \cos +1=0 $$ and $$ \sin \alpha0 $$ a) $$ \sqrt{3} \cdot \sin \alpha-\frac{\sqrt{2}}{\cos \alpha} $$ a) $$ \frac{1}{\cos ^{3} x}+ an ^{3}-x $$

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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) $$ 	an \alpha=\sin \alpha $$ b) $$ 2 \sin ^{2}-\cos ^{2}= $$ (2) $$ \sqrt{2} 	an x+1=0 $$ and $$ \cos 0 x>0 $$ a) $$ \sqrt{3} \cdot \sin \alpha-\frac{\sqrt{2}}{\cos \alpha} $$ a) $$ \frac{1}{\cos ^{3} x}+	an ^{3}-x $$

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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 $$ an \alpha $$: $$ an \alpha = \frac{\sin \alpha}{\cos \alpha} $$ 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 $$ \cos \alpha = -\frac{1}{2} $$: $$ \sin^2 \alpha + \left(-\frac{1}{2} ight)^2 = 1 $$ $$ \sin^2 \alpha + \frac{1}{4} = 1 $$ $$ \sin^2 \alpha = 1 - \frac{1}{4} $$ $$ \sin^2 \alpha = \frac{3}{4} $$ Since $$ \sin \alpha < 0 $$, we take the negative square root: $$ \sin \alpha = -\frac{\sqrt{3}}{2} $$ Now, calculate $$ an \alpha $$: $$ an \alpha = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} $$ So, $$ an \alpha = \sqrt{3} $$. Now, let's evaluate the expressions: (a) $$ an \alpha = \sin \alpha $$ We have $$ an \alpha = \sqrt{3} $$ and $$ \sin \alpha = -\frac{\sqrt{3}}{2} $$. Clearly, $$ \sqrt{3} eq -\frac{\sqrt{3}}{2} $$, 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: $$ 2 imes \frac{3}{4} - \frac{1}{4} = \frac{6}{4} - \frac{1}{4} = \frac{5}{4} $$ So, $$ 2 \sin^2 \alpha - \cos^2 \alpha = \frac{5}{4} $$. Now, moving on to the second set of equations: 2) $$ \sqrt{2} an x +1=0 $$ and $$ \cos x > 0 $$ Let's solve for $$ an x $$: $$ \sqrt{2} an x = -1 $$ $$ an x = -\frac{1}{\sqrt{2}} $$ Now, since $$ \cos x > 0 $$, and $$ an x = \frac{\sin x}{\cos x} $$, this implies that $$ \sin x $$ must be negative because $$ an x $$ 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 $$ \sin^2 x + \cos^2 x = 1 $$: $$ \sin^2 x + \left(\frac{1}{\sqrt{2}} ight)^2 = 1 $$ $$ \sin^2 x + \frac{1}{2} = 1 $$ $$ \sin^2 x = 1 - \frac{1}{2} = \frac{1}{2} $$ 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: $$ \sqrt{3} imes \left(-\frac{1}{\sqrt{2}} ight) - \frac{\sqrt{2}}{\frac{1}{\sqrt{2}}} = -\frac{\sqrt{3}}{\sqrt{2}} - 2 = -\frac{\sqrt{6}}{2} - 2 $$ So, $$ \sqrt{3} \cdot \sin \alpha - \frac{\sqrt{2}}{\cos \alpha} = -\frac{\sqrt{6}}{2} - 2 $$ (b) $$ \frac{1}{\cos^3 x} + an^3 x $$ Plugging in the values: $$ \frac{1}{\left(\frac{1}{\sqrt{2}} ight)^3} + \left(-\frac{1}{\sqrt{2}} ight)^3 = \frac{1}{\frac{1}{2\sqrt{2}}} + \left(-\frac{1}{2\sqrt{2}} ight) = 2\sqrt{2} - \frac{1}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{3}{2\sqrt{2}} $$ So, $$ \frac{1}{\cos^3 x} + an^3 x = \frac{3}{2\sqrt{2}} $$ 1) For the first set of equations: (a) $$ an \alpha = \sin \alpha $$ 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) $$ \frac{1}{\cos^3 x} + an^3 x = \frac{3}{2\sqrt{2}} $$.

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