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Answer:
617 ^ 15
Step-by-step explanation:
If the bases are the same when multiplying, add the exponents
617 ^ 9 * 617^6
617 ^(9+6)
617 ^ 15
Answer:
![\large\boxed{x=0\ and\ x=\pi}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7Bx%3D0%5C%20and%5C%20x%3D%5Cpi%7D)
Step-by-step explanation:
![\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2](https://tex.z-dn.net/?f=%5Ctan%5E2x%5Csec%5E2x%2B2%5Csec%5E2x-%5Ctan%5E2x%3D2%5C%5C%5C%5C%5Ctext%7BUse%7D%5C%20%5Ctan%20x%3D%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%2C%5C%20%5Csec%20x%3D%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%3A%5C%5C%5C%5C%5Cleft%28%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%2B2%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%20x%7D%5Cright%29%5E2-%5Cleft%28%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%5Cright%29%5E2%3D2%5C%5C%5C%5C%5Cleft%28%5Cdfrac%7B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%5Cright%29%5Cleft%28%5Cdfrac%7B1%7D%7B%5Ccos%5E2x%7D%5Cright%29%2B%5Cdfrac%7B2%7D%7B%5Ccos%5E2x%7D-%5Cdfrac%7B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D2)
![\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csin%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Ctext%7BUse%7D%5C%20%5Csin%5E2x%2B%5Ccos%5E2x%3D1%5Cto%5Csin%5E2x%3D1-%5Ccos%5E2x%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-%281-%5Ccos%5E2x%29%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B2-1%2B%5Ccos%5E2x%7D%7B%5Ccos%5E2x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B1%2B%5Ccos%5E2x%7D%7B%5Ccos%5E2x%7D%3D2)
![\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x](https://tex.z-dn.net/?f=%5Cdfrac%7B1-%5Ccos%5E2x%7D%7B%28%5Ccos%5E2x%29%5E2%7D%2B%5Cdfrac%7B%281%2B%5Ccos%5E2x%29%28%5Ccos%5E2x%29%7D%7B%28%5Ccos%5E2x%29%5E2%7D%3D2%5Cqquad%5Ctext%7BUse%20the%20distributive%20property%7D%5C%5C%5C%5C%5Cdfrac%7B1-%5Ccos%5E2x%2B%5Ccos%5E2x%2B%5Ccos%5E4x%7D%7B%5Ccos%5E4x%7D%3D2%5C%5C%5C%5C%5Cdfrac%7B1%2B%5Ccos%5E4x%7D%7B%5Ccos%5E4x%7D%3D2%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%7D%5C%20%5Ccos%5E4x%5Cneq0%5C%5C%5C%5C1%2B%5Ccos%5E4x%3D2%5Ccos%5E4x%5Cqquad%5Ctext%7Bsubtract%7D%5C%20%5Ccos%5E4x%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5C1%3D%5Ccos%5E4x%5Ciff%20%5Ccos%20x%3D%5Cpm%5Csqrt1%5Cto%5Ccos%20x%3D%5Cpm1%5C%5C%5C%5C%20x%3Dk%5Cpi%5C%20for%5C%20k%5Cin%5Cmathbb%7BZ%7D%5C%5C%5C%5C%5Ctext%7BOn%20the%20interval%7D%5C%200%5Cleq%20x%3C2%5Cpi%2C%5C%20%5Ctext%7Bthe%20solutions%20are%7D%5C%20x%3D0%5C%20%5Ctext%7Band%7D%5C%20x%3D%5Cpi.)
OPTION C is the correct answer.
Hey there!
The answer for your problem is: a line is a set of 2 or more points in a plane
a point is a specific location in a plane
Let me know if you'd like me to explain how I got this answer!