Instead of setting t = sin(x), first raise both sides by 2 and set t=3-sinx. After some manipulation you can get (t*ln2) e^(t*ln2) = (2ln2)e^(2ln2) where you can apply the product log to get t = 2, or sin(x) = 1.
We raise Both Sides to second and we get 2^sinx=3-sinx. Let sinx=t , -1 =or< t < or=1. So 2^ t or= 2 (2). Hence from (1) and (2) the Only Case is Both are equal with 2. So 2^t=2=2^1 so t=1 so sinx=1=sinπ/2 so x=2kπ+π/2 or x=2kπ+π-π/2=2kπ+π/2, k=Integer. SAME as First. Finally there is One Solution x=2kπ+π/2., k=Any Integer.
Answer 90 degrees or 1.578 radian and 90 degrees + 360 degrees (n) or 90 degrees + 2pi or 7.854 radian, in which n = 1, 2, 3, 4, 5... 90 degrees - 360 degrees or 90 degrees - 2 pi = - 270 degrees, -630 degrees. Though sin -270, =1 and sine -630 =1 and sin -990 = 1 and so on, you cant get the log of a negative so the negative numbers are out. So i remove the negative. Log 2 (3 - sinx) = sin x 2^ sinx = 3- sinx change to exponential form 2^ sinx + sinx = 2 + 1 2^ sinx + sinx = 2^1 + 1 Hence, sinx = 1 the sine of 90 degrees = 1, and sine of 90 + 360 or 450 degrees , and ( 450 + 360 ) degrees = 1 etc Also, the sine of 90 - 360 degrees = 1 or -270, and -270- 360, and -630 - 360 degrees Answer 90 degrees + 2 pi (N) and 90 degrees in which n = 1, 2 ,3, 4, 5,etc
Solved this problem in my head. You just look at the range of values of both parts and see only one intersection point. sinx = 1.
2
Instead of setting t = sin(x), first raise both sides by 2 and set t=3-sinx. After some manipulation you can get (t*ln2) e^(t*ln2) = (2ln2)e^(2ln2) where you can apply the product log to get t = 2, or sin(x) = 1.
We raise Both Sides to second and we get 2^sinx=3-sinx. Let sinx=t , -1 =or< t < or=1. So 2^ t or= 2 (2). Hence from (1) and (2) the Only Case is Both are equal with 2. So 2^t=2=2^1 so t=1 so sinx=1=sinπ/2 so x=2kπ+π/2 or x=2kπ+π-π/2=2kπ+π/2, k=Integer. SAME as First. Finally there is One Solution x=2kπ+π/2., k=Any Integer.
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Got it!
log2(3-sinx)=sinx
Let u=sinx
The given equation transforms to log2(3-u)=u => 3-u=2^u (1)
Let f(x)=2^x+x-3, where abs(x)
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log2 （3-sinx）>=logz2 (3-1)=1
sinx
pi/2 +k2pi
sinx = 1
X=90°
Answer 90 degrees or 1.578 radian and 90 degrees + 360 degrees (n) or 90 degrees + 2pi or 7.854 radian, in which n = 1, 2, 3, 4, 5...
90 degrees - 360 degrees or 90 degrees - 2 pi = - 270 degrees, -630 degrees.
Though sin -270, =1 and sine -630 =1 and sin -990 = 1 and so on, you cant get the log of a negative so the negative numbers are out. So i remove the negative.
Log 2 (3 - sinx) = sin x
2^ sinx = 3- sinx change to exponential form
2^ sinx + sinx = 2 + 1
2^ sinx + sinx = 2^1 + 1
Hence, sinx = 1
the sine of 90 degrees = 1, and sine of 90 + 360 or 450 degrees , and ( 450 + 360 ) degrees = 1 etc
Also, the sine of 90 - 360 degrees = 1 or -270, and -270- 360, and -630 - 360 degrees
Answer 90 degrees + 2 pi (N)
and 90 degrees
in which n = 1, 2 ,3, 4, 5,etc
Why do u need absolute value of sinx?
To express the compound inequality
log2 (3-sinx)=sinx
log2 (3-sin90)=sin90
log2(3-1)=11=1x=90