Answer:
- θ = (2/5)πk or π(k +1/2) . . . . . for any integer k
Step-by-step explanation:
We can make use of the identities ...
[tex]\cos{\alpha}-\cos{\beta}=-2\sin{\dfrac{\alpha+\beta}{2}}\sin{\dfrac{\alpha-\beta}{2}}\\\\\sin{\alpha}+\sin{\beta}=2\sin{\dfrac{\alpha+\beta}{2}}\cos{\dfrac{\alpha-\beta}{2}}[/tex]
These let us rewrite the equation as ...
[tex]0=\cos{\theta}-\cos{2\theta}+\cos{3\theta}-\cos{4\theta}\\\\0=-2\sin{\dfrac{\theta+2\theta}{2}}\sin{\dfrac{\theta-2\theta}{2}}-2\sin{\dfrac{3\theta+4\theta}{2}}\sin{\dfrac{3\theta-4\theta}{2}}\\\\0=2\sin{\dfrac{\theta}{2}}\left(\sin{\dfrac{3\theta}{2}}+\sin{\dfrac{7\theta}{2}}\right)\\\\0=4\sin{\dfrac{\theta}{2}}\sin{\dfrac{3\theta+7\theta}{4}}\cos{\dfrac{3\theta-7\theta}{4}}\\\\0=4\sin{\dfrac{\theta}{2}}\sin{\dfrac{5\theta}{2}}\cos{\theta}[/tex]
The solutions are the values of θ that make the factors zero. That is, ...
θ = 2πk . . . . for any integer k
θ = (2/5)πk . . . . for any integer k (includes the above cases)
θ = π(k +1/2) . . . . for any integer k
