+1911 Let x be an acute angle such that \tan x + \sec x = 0. Find $\cos x$.
+399 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.\(\)
+399 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.\(\)
