In[8]:=
Clear@"Global`*"D; $Line = 0;
In[1]:=
Sin@3 xD + Cos@2 xD - Sin@xD Š 0;
Replace Sin[3x] and Cos[2x] in In[1] above with 3 Sin@xD - 4 Sin@xD3 and 1 - 2 Sin@xD2 respectively.
Simplify the result.
In[2]:=
Out[2]=
%1 . 9 Sin@3 xD ® I3 Sin@xD - 4 Sin@xD3 M, Cos@2 xD ® I1 - 2 Sin@xD2 M=
1 + 2 Sin@xD - 2 Sin@xD2 - 4 Sin@xD3 Š 0
Solve the above equation for x. %2 refers to Out[2].
In[3]:=
Solve@%2, xD
Solve::ifun : Inverse functions are being used by Solve, so
some solutions may not be found; use Reduce for complete solution information. ‡
Π
Out[3]=
::x ® -
Π
>, :x ® -
4
Π
>, :x ®
6
>>
4
Out[2] can aslo be factored.
In[4]:=
Out[4]=
%2  Factor
- H1 + 2 Sin@xDL I- 1 + 2 Sin@xD2 M Š 0
Verify the answers using the orginal equation. True means that the RHS and the LHS are equal after post substitution evaluation.
In[5]:=
Out[5]=
%1 . %3
8True, True, True<
Run Reduce per the recommendation of the In[3] alert. x is the independent variable and the domain is limited to
real numbers.
In[6]:=
ReduceA1 + 2 Sin@xD - 2 Sin@xD2 - 4 Sin@xD3 Š 0, x, RealsE
7Π
Π
Out[6]=
C@1D Î Integers && x Š -
+ 2 Π C@1D ÈÈ x Š
6
5Π
Π
xŠ-
+ 2 Π C@1D ÈÈ
6
+ 2 Π C@1D ÈÈ x Š
4
3Π
Π
+ 2 Π C@1D ÈÈ x Š
4
+ 2 Π C@1D ÈÈ x Š
4
Evaluate the above with C[1] set to zero.
In[7]:=
%6 . C@1D ® 0
7Π
Π
Out[7]=
xŠ-
ÈÈ x Š
6
6
5Π
Π
ÈÈ x Š -
ÈÈ x Š
4
4
3Π
Π
ÈÈ x Š
ÈÈ x Š
4
+ 2 Π C@1D
4
4