Question 12:
Step 1: Rewrite the equation.
The given equation is:
sin(2θ−30∘)−cos(32θ)=0
Rearrange the equation:
sin(2θ−30∘)=cos(32θ)
Step 2: Use the trigonometric identity cosx=sin(90∘−x).
Apply this identity to the right side of the equation:
cos(32θ)=sin(90∘−32θ)
Substitute this back into the equation:
sin(2θ−30∘)=sin(90∘−32θ)
Step 3: Solve for θ using the general solutions for sinA=sinB.
The general solutions for sinA=sinB are A=B+360∘n or A=(180∘−B)+360∘n, where n is an integer.
Case 1: 2θ−30∘=90∘−32θ+360∘n
Combine terms with θ:
2θ+32θ=90∘+30∘+360∘n
36θ+2θ=120∘+360∘n
38θ=120∘+360∘n
Multiply by 3:
8θ=3(120∘+360∘n)
8θ=360∘+1080∘n
Divide by 8:
θ=8360∘+81080∘n
θ=45∘+135∘n
For n=0, θ=45∘.
For n=1, θ=45∘+135∘=180∘.
For n=2, θ=45∘+270∘=315∘.
Case 2: 2θ−30∘=(180∘−(90∘−32θ))+360∘n
2θ−30∘=180∘−90∘+32θ+360∘n
2θ−30∘=90∘+32θ+360∘n
Combine terms with θ:
2θ−32θ=90∘+30∘+360∘n
36θ−2θ=120∘+360∘n
34θ=120∘+360∘n
Multiply by 3:
4θ=3(120∘+360∘n)
4θ=360∘+1080∘n
Divide by 4:
θ=4360∘+41080∘n
θ=90∘+270∘n
For n=0, θ=90∘.
Assuming the solutions are required in the range 0∘≤θ<360∘, the values for θ are:
$\boxed{\theta = 45^\circ, 90^\circ, 180^\circ, 315
