Mixing
If $cos^6 (x) + sin^4 (x)=1$, find $x$ if $xin [0, dfrac {pi}{2}]$
$begingroup$
If $cos^6 (x) + sin^4 (x)=1$, find $x$ if $xin [0, dfrac {pi}{2}]$
My attempt:
$$cos^6 (x) + sin^4 (x)=1$$
$$cos^6 (x) + (1-cos^2 (x))^{2}=1$$
$$cos^6 (x) + 1 - 2cos^2 (x) + cos^4 (x) = 1$$
$$cos^6 (x) + cos^4 (x) - 2cos^2 (x)=0$$
trigonometry
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
$endgroup$
add a comment |
$begingroup$
If $cos^6 (x) + sin^4 (x)=1$, find $x$ if $xin [0, dfrac {pi}{2}]$
My attempt:
$$cos^6 (x) + sin^4 (x)=1$$
$$cos^6 (x) + (1-cos^2 (x))^{2}=1$$
$$cos^6 (x) + 1 - 2cos^2 (x) + cos^4 (x) = 1$$
$$cos^6 (x) + cos^4 (x) - 2cos^2 (x)=0$$
trigonometry
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
$endgroup$
2
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34
add a comment |
$begingroup$
If $cos^6 (x) + sin^4 (x)=1$, find $x$ if $xin [0, dfrac {pi}{2}]$
My attempt:
$$cos^6 (x) + sin^4 (x)=1$$
$$cos^6 (x) + (1-cos^2 (x))^{2}=1$$
$$cos^6 (x) + 1 - 2cos^2 (x) + cos^4 (x) = 1$$
$$cos^6 (x) + cos^4 (x) - 2cos^2 (x)=0$$
trigonometry
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
$endgroup$
If $cos^6 (x) + sin^4 (x)=1$, find $x$ if $xin [0, dfrac {pi}{2}]$
My attempt:
$$cos^6 (x) + sin^4 (x)=1$$
$$cos^6 (x) + (1-cos^2 (x))^{2}=1$$
$$cos^6 (x) + 1 - 2cos^2 (x) + cos^4 (x) = 1$$
$$cos^6 (x) + cos^4 (x) - 2cos^2 (x)=0$$
trigonometry
trigonometry
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
asked Jan 25 at 15:57
blue_eyed_...blue_eyed_...
3,29521752
3,29521752
2
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34
add a comment |
2
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34
2
2
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint:
Clearly, $sin x=0iffcos^2x=?,$
$cos x=0iffsin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
Hint If we rewrite $u = cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $cos^6 x leq cos^2 x$ and $sin^4 x leq sin^2 x$, and in both cases equality holds (for $x in [0, frac{pi}{2}]$) only for $x = 0, frac{pi}{2}$.
answered Jan 25 at 16:05
TravisTravis
63.4k769150
$endgroup$
add a comment |
$begingroup$
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 Rightarrow y(y^2+y-2) = 0 Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 leq y leq 1$:
$$y = 0, 1 Rightarrow cos^2(x) = 0, 1 Rightarrow x = 0, frac{pi}{2},$$
for $x in [0, frac{pi}{2}]$.
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
Let $cos x = c, sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $sin x = 0$ or $cos x = 0$.
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
$begingroup$
$$1=cos^6x+sin^4xleqcos^2x+sin^2x=1,$$ which gives the following system.
$$cos^6x=cos^2x$$ and $$sin^4x=sin^2x.$$
Can you end it now?
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Clearly, $sin x=0iffcos^2x=?,$
$cos x=0iffsin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Clearly, $sin x=0iffcos^2x=?,$
$cos x=0iffsin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Clearly, $sin x=0iffcos^2x=?,$
$cos x=0iffsin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
Hint:
Clearly, $sin x=0iffcos^2x=?,$
$cos x=0iffsin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 16:03
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
add a comment |
add a comment |
$begingroup$
Hint If we rewrite $u = cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $cos^6 x leq cos^2 x$ and $sin^4 x leq sin^2 x$, and in both cases equality holds (for $x in [0, frac{pi}{2}]$) only for $x = 0, frac{pi}{2}$.
answered Jan 25 at 16:05
TravisTravis
63.4k769150
$endgroup$
add a comment |
$begingroup$
Hint If we rewrite $u = cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $cos^6 x leq cos^2 x$ and $sin^4 x leq sin^2 x$, and in both cases equality holds (for $x in [0, frac{pi}{2}]$) only for $x = 0, frac{pi}{2}$.
answered Jan 25 at 16:05
TravisTravis
63.4k769150
$endgroup$
add a comment |
$begingroup$
Hint If we rewrite $u = cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $cos^6 x leq cos^2 x$ and $sin^4 x leq sin^2 x$, and in both cases equality holds (for $x in [0, frac{pi}{2}]$) only for $x = 0, frac{pi}{2}$.
answered Jan 25 at 16:05
TravisTravis
63.4k769150
$endgroup$
Hint If we rewrite $u = cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $cos^6 x leq cos^2 x$ and $sin^4 x leq sin^2 x$, and in both cases equality holds (for $x in [0, frac{pi}{2}]$) only for $x = 0, frac{pi}{2}$.
answered Jan 25 at 16:05
TravisTravis
63.4k769150
answered Jan 25 at 16:05
TravisTravis
63.4k769150
answered Jan 25 at 16:05
TravisTravis
63.4k769150
answered Jan 25 at 16:05
TravisTravis
63.4k769150
63.4k769150
add a comment |
add a comment |
$begingroup$
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 Rightarrow y(y^2+y-2) = 0 Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 leq y leq 1$:
$$y = 0, 1 Rightarrow cos^2(x) = 0, 1 Rightarrow x = 0, frac{pi}{2},$$
for $x in [0, frac{pi}{2}]$.
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 Rightarrow y(y^2+y-2) = 0 Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 leq y leq 1$:
$$y = 0, 1 Rightarrow cos^2(x) = 0, 1 Rightarrow x = 0, frac{pi}{2},$$
for $x in [0, frac{pi}{2}]$.
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 Rightarrow y(y^2+y-2) = 0 Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 leq y leq 1$:
$$y = 0, 1 Rightarrow cos^2(x) = 0, 1 Rightarrow x = 0, frac{pi}{2},$$
for $x in [0, frac{pi}{2}]$.
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
$endgroup$
Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 Rightarrow y(y^2+y-2) = 0 Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 leq y leq 1$:
$$y = 0, 1 Rightarrow cos^2(x) = 0, 1 Rightarrow x = 0, frac{pi}{2},$$
for $x in [0, frac{pi}{2}]$.
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
edited Jan 25 at 17:16
J. W. Tanner
3,4601320
3,4601320
answered Jan 25 at 16:02
OmGOmG
2,512824
answered Jan 25 at 16:02
OmGOmG
2,512824
answered Jan 25 at 16:02
OmGOmG
2,512824
2,512824
add a comment |
add a comment |
$begingroup$
Let $cos x = c, sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $sin x = 0$ or $cos x = 0$.
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
$begingroup$
Let $cos x = c, sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $sin x = 0$ or $cos x = 0$.
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
$begingroup$
Let $cos x = c, sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $sin x = 0$ or $cos x = 0$.
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
$endgroup$
Let $cos x = c, sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $sin x = 0$ or $cos x = 0$.
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
answered Jan 25 at 16:07
DeepakDeepak
17.5k11539
17.5k11539
add a comment |
add a comment |
$begingroup$
$$1=cos^6x+sin^4xleqcos^2x+sin^2x=1,$$ which gives the following system.
$$cos^6x=cos^2x$$ and $$sin^4x=sin^2x.$$
Can you end it now?
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$1=cos^6x+sin^4xleqcos^2x+sin^2x=1,$$ which gives the following system.
$$cos^6x=cos^2x$$ and $$sin^4x=sin^2x.$$
Can you end it now?
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$1=cos^6x+sin^4xleqcos^2x+sin^2x=1,$$ which gives the following system.
$$cos^6x=cos^2x$$ and $$sin^4x=sin^2x.$$
Can you end it now?
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$$1=cos^6x+sin^4xleqcos^2x+sin^2x=1,$$ which gives the following system.
$$cos^6x=cos^2x$$ and $$sin^4x=sin^2x.$$
Can you end it now?
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 25 at 16:47
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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2
$begingroup$
Try letting $t=cos^2(x)$ then the equation becomes $t^3+t^2-2t=0$ and we can factorise $ t $ to solve a quadratic.
$endgroup$
– Peter Foreman
Jan 25 at 16:01
$begingroup$
You recieved 6 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:34