How do I solve this inequality? $3sin(2x)> sin(x)+cos(x)+1$
$begingroup$
How do I solve this inequality?
$3sin(2x)> sin(x)+cos(x)+1$
I am unable to think of a method to solve this.
Writing $sin(2x)=2sin(x)cos(x)$ don't seem to help
trigonometry inequality
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
$endgroup$
add a comment |
$begingroup$
How do I solve this inequality?
$3sin(2x)> sin(x)+cos(x)+1$
I am unable to think of a method to solve this.
Writing $sin(2x)=2sin(x)cos(x)$ don't seem to help
trigonometry inequality
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
$endgroup$
add a comment |
$begingroup$
How do I solve this inequality?
$3sin(2x)> sin(x)+cos(x)+1$
I am unable to think of a method to solve this.
Writing $sin(2x)=2sin(x)cos(x)$ don't seem to help
trigonometry inequality
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
$endgroup$
How do I solve this inequality?
$3sin(2x)> sin(x)+cos(x)+1$
I am unable to think of a method to solve this.
Writing $sin(2x)=2sin(x)cos(x)$ don't seem to help
trigonometry inequality
trigonometry inequality
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
asked Dec 26 '18 at 18:00
AbhayAbhay
3219
3219
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint
$$sin x+cos x=u,u^2=1+sin2x$$
$$3(u^2-1)>u+1$$
$$iff3u^2-u-4>0$$
$$iff(3u-4)(u+1)>0$$
$implies$ either $u>$max$left(-1,dfrac43right)=?$
or $u<$min$left(-1,dfrac43right)=?$
Now keep in mind $sin x+cos x=sqrt2sinleft(x+dfracpi4right)implies-sqrt2le ulesqrt2$
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint
$$sin x+cos x=u,u^2=1+sin2x$$
$$3(u^2-1)>u+1$$
$$iff3u^2-u-4>0$$
$$iff(3u-4)(u+1)>0$$
$implies$ either $u>$max$left(-1,dfrac43right)=?$
or $u<$min$left(-1,dfrac43right)=?$
Now keep in mind $sin x+cos x=sqrt2sinleft(x+dfracpi4right)implies-sqrt2le ulesqrt2$
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
add a comment |
$begingroup$
Hint
$$sin x+cos x=u,u^2=1+sin2x$$
$$3(u^2-1)>u+1$$
$$iff3u^2-u-4>0$$
$$iff(3u-4)(u+1)>0$$
$implies$ either $u>$max$left(-1,dfrac43right)=?$
or $u<$min$left(-1,dfrac43right)=?$
Now keep in mind $sin x+cos x=sqrt2sinleft(x+dfracpi4right)implies-sqrt2le ulesqrt2$
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
add a comment |
$begingroup$
Hint
$$sin x+cos x=u,u^2=1+sin2x$$
$$3(u^2-1)>u+1$$
$$iff3u^2-u-4>0$$
$$iff(3u-4)(u+1)>0$$
$implies$ either $u>$max$left(-1,dfrac43right)=?$
or $u<$min$left(-1,dfrac43right)=?$
Now keep in mind $sin x+cos x=sqrt2sinleft(x+dfracpi4right)implies-sqrt2le ulesqrt2$
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Hint
$$sin x+cos x=u,u^2=1+sin2x$$
$$3(u^2-1)>u+1$$
$$iff3u^2-u-4>0$$
$$iff(3u-4)(u+1)>0$$
$implies$ either $u>$max$left(-1,dfrac43right)=?$
or $u<$min$left(-1,dfrac43right)=?$
Now keep in mind $sin x+cos x=sqrt2sinleft(x+dfracpi4right)implies-sqrt2le ulesqrt2$
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
edited Dec 27 '18 at 6:40
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 26 '18 at 18:05
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
add a comment |
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
2
2
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
$begingroup$
Have to remember this one. :)
$endgroup$
– greedoid
Dec 26 '18 at 18:10
add a comment |
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