Answer:
[tex]\displaystyle \sec x=\pm\sqrt{2}[/tex]
Step-by-step explanation:
Trigonometric Equations
We need to recall the identity:
[tex]\sin^2 x+\cos ^2 x=1[/tex]
Solving for the sine:
[tex]\sin^2 x=1-\cos ^2 \qquad\qquad [1][/tex]
We are given the equation:
[tex]\sin x-\cos x=0[/tex]
Or, equivalently:
[tex]\sin x=\cos x[/tex]
Squaring:
[tex]\sin^2 x=\cos^2 x[/tex]
Substituting from [1]
[tex]1-\cos ^2 =\cos^2 x[/tex]
Simplifying:
[tex]2\cos ^2=1[/tex]
Solving:
[tex]\displaystyle \cos^2x=\frac{1}{2}[/tex]
[tex]\displaystyle \cos x=\pm\sqrt{\frac{1}{2}}[/tex]
Since:
[tex]\sec x = \frac{1}{\cos x}:[/tex]
[tex]\mathbf{\displaystyle \sec x=\pm\sqrt{2}}[/tex]
