If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555——————————
Solve the trigonometric equation:
![\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Csqrt%7B2%5C%2Ctan%5C%2Cx%5C%2Ccos%5C%2Cx%7D-tan%5C%2Cx%3D0%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Csqrt%7B2%5Ccdot%20%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5C%2Cx%7D%5Ccdot%20cos%5C%2Cx%7D-tan%5C%2Cx%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Csqrt%7B2%5Ccdot%20sin%5C%2Cx%7D%3Dtan%5C%2Cx%5Cqquad%5Cquad%28i%29%7D)
Restriction for the solution:
![\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Cmathsf%7Bsin%5C%2Cx%5Cge%200%7D%5C%5C%5C%5C%20%5Cmathsf%7Btan%5C%2Cx%5Cge%200%7D%20%5Cend%7Barray%7D%20%5Cright.)
Square both sides of
(i):
![\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}](https://tex.z-dn.net/?f=%5Cmathsf%7B%28%5Csqrt%7B2%5Ccdot%20sin%5C%2Cx%7D%29%5E2%3D%28tan%5C%2Cx%29%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B2%5Ccdot%20sin%5C%2Cx%3Dtan%5E2%5C%2Cx%7D%5C%5C%5C%5C%20%5Cmathsf%7B2%5Ccdot%20sin%5C%2Cx-tan%5E2%5C%2Cx%3D0%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7B2%5Ccdot%20sin%5C%2Cx%5Ccdot%20cos%5E2%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D-%5Cdfrac%7Bsin%5E2%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%282%5C%2Ccos%5E2%5C%2Cx-sin%5C%2Cx%20%5Cright%20%29%3D0%5Cqquad%5Cquad%20but~~cos%5E2%20x%3D1-sin%5E2%20x%7D)
![\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5Ccdot%20%281-sin%5E2%5C%2Cx%29-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2-2%5C%2Csin%5E2%5C%2Cx-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B-%5C%2C%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2Cx%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D)
Let
![\mathsf{sin\,x=t\qquad (0\le t](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2Cx%3Dt%5Cqquad%20%280%5Cle%20t%3C1%29%7D)
So the equation becomes
![\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}](https://tex.z-dn.net/?f=%5Cmathsf%7Bt%5Ccdot%20%282t%5E2%2Bt-2%29%3D0%5Cqquad%5Cquad%20%28ii%29%7D%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Brcl%7D%20%5Cmathsf%7Bt%3D0%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7B2t%5E2%2Bt-2%3D0%7D%20%5Cend%7Barray%7D)
Solving the quadratic equation:
![\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.](https://tex.z-dn.net/?f=%5Cmathsf%7B2t%5E2%2Bt-2%3D0%7D%5Cquad%5Clongrightarrow%5Cquad%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Cmathsf%7Ba%3D2%7D%5C%5C%20%5Cmathsf%7Bb%3D1%7D%5C%5C%20%5Cmathsf%7Bc%3D-2%7D%20%5Cend%7Barray%7D%20%5Cright.)
![\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5CDelta%3Db%5E2-4ac%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5CDelta%3D1%5E2-4%5Ccdot%202%5Ccdot%20%28-2%29%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5CDelta%3D1%2B16%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5CDelta%3D17%7D)
![\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}](https://tex.z-dn.net/?f=%5Cmathsf%7Bt%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bt%3D%5Cdfrac%7B-1%5Cpm%5Csqrt%7B17%7D%7D%7B2%5Ccdot%202%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bt%3D%5Cdfrac%7B-1%5Cpm%5Csqrt%7B17%7D%7D%7B4%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Brcl%7D%20%5Cmathsf%7Bt%3D%5Cdfrac%7B-1%2B%5Csqrt%7B17%7D%7D%7B4%7D%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7Bt%3D%5Cdfrac%7B-1-%5Csqrt%7B17%7D%7D%7B4%7D%7D%20%5Cend%7Barray%7D)
You can discard the negative value for
t. So the solution for
(ii) is
![\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%20%5Cmathsf%7Bt%3D0%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7Bt%3D%5Cdfrac%7B%5Csqrt%7B17%7D-1%7D%7B4%7D%7D%20%5Cend%7Barray%7D)
Substitute back for
t = sin x. Remember the restriction for
x:
![\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%20%5Cmathsf%7Bsin%5C%2Cx%3D0%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7Bsin%5C%2Cx%3D%5Cdfrac%7B%5Csqrt%7B17%7D-1%7D%7B4%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bx%3D0%2Bk%5Ccdot%20180%5E%5Ccirc%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7Bx%3Darcsin%5Cbigg%28%5Cdfrac%7B%5Csqrt%7B17%7D-1%7D%7B4%7D%5Cbigg%29%2Bk%5Ccdot%20360%5E%5Ccirc%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bx%3Dk%5Ccdot%20180%5E%5Ccirc%7D%26%5Ctextsf%7B%20or%20%7D%26%5Cmathsf%7Bx%3D51.33%5E%5Ccirc%20%2Bk%5Ccdot%20360%5E%5Ccirc%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7Bsolution.%7D%20%5Cend%7Barray%7D)
where
k is an integer.
I hope this helps. =)
Answer:
12
Step-by-step explanation:
∛1728 = ∛2*2*2*2*2*2*3*3*3*3 = 2*2*3 = 12
Answer:
12
Step-by-step explanation:
Let's put the equations in standard form. For the first equation, we have:
−11y=6(z+1)-13y
2y−6z=6
y−3z=3
The second equation is:
4y−24=c(z−1)
4y−cz=24−c
If we multiply the first equation by 4, we get:
4y-12z=12
Comparing the two equations, we see that if c=12, both equations will be the same and there will be infinitely many solutions.
The correct value of c is 12.