7.5 Solving trigonometric equations  (Page 4/7)

Using grouping, this quadratic can be factored. Either make the real substitution, $\sin \theta =u,$ or imagine it, as we factor:

Now set each factor equal to zero.

Next solve for $\theta :\sin \theta \ne \frac{3}{2},$ as the range of the sine function is $\left[\mathrm{-1},1\right].$ However, $\sin \theta =1,$ giving the solution $\frac{\pi }{2}.$

This problem should appear familiar as it is similar to a quadratic. Let $\sin \theta =x.$ The equation becomes $2x^2+x=0.$ We begin by factoring:

Set each factor equal to zero.

Then, substitute back into the equation the original expression $\sin \theta$ for $x.$ Thus,

The solutions within the domain $0\le \theta <2\pi$ are $0,\pi ,\frac{7\pi }{6},\frac{11\pi }{6}.$

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

We can factor using grouping. Solution values of $\theta$ can be found on the unit circle:

Notice that the left side of the equation is the difference formula for cosine.

From the unit circle in [link] , we see that $\cos x=\frac{\sqrt{3}}{2}$ when $x=\frac{\pi }{6},\frac{11\pi }{6}.$

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

So, if $\cos \theta =-\frac{1}{2},$ then $\theta =\frac{2\pi }{3}\pm 2\pi k$ and $\theta =\frac{4\pi }{3}\pm 2\pi k;$ if $\cos \theta =1,$ then $\theta =0\pm 2\pi k.$