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# Trig Equation

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Let x be an acute angle such that \tan x + \sec x = 0. Find $\cos x$.

Feb 11, 2024

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What do we know about tan(x) and sec(x)? We know $$\tan(x)=\frac{\sin(x)}{\cos(x)}, sec(x)=\frac{1}{cos(x)}$$. Now use these, and replace them in the equation.

We get $$\frac{sin(x)}{\cos(x)}+\frac{1}{cos(x)}=0$$. Multiplying by cos(x), with the restriction, cos(x) is not equal to 0, gives us $$\sin(x)+1=0$$, or $$sin(x)=-1$$. So, now we use the formula, $${\sin}^{2}(x)+{\cos}^{2}(x)=1$$, we get $$cos^2(x)=0, cos(x)=0$$. Notice there is only one solution, because +/- 0 is always 0.

But we see from ealier that cos(x) is not 0, so we have no solution.

Feb 11, 2024

#1
+394
+2
What do we know about tan(x) and sec(x)? We know $$\tan(x)=\frac{\sin(x)}{\cos(x)}, sec(x)=\frac{1}{cos(x)}$$. Now use these, and replace them in the equation.
We get $$\frac{sin(x)}{\cos(x)}+\frac{1}{cos(x)}=0$$. Multiplying by cos(x), with the restriction, cos(x) is not equal to 0, gives us $$\sin(x)+1=0$$, or $$sin(x)=-1$$. So, now we use the formula, $${\sin}^{2}(x)+{\cos}^{2}(x)=1$$, we get $$cos^2(x)=0, cos(x)=0$$. Notice there is only one solution, because +/- 0 is always 0.
But we see from ealier that cos(x) is not 0, so we have no solution.