Solve Trigonometric Equations

Example on how to solve a trigonometric equation corresponding to question 7 in trigonometry_2.

Example 1


What is the solution set for the equation $-\cos ^2 x=-2(\sin x+1)$ for $0 < x < 2\pi$ ?

  1. $\{\pi \}$
  2. $\{0, \pi \}$
  3. $\{0, \pi , 2\pi \}$
  4. $\{0, \dfrac{\pi}{2} , 2\pi \}$
  5. $\{ \dfrac{3\pi}{2} \}$

Solution


  1. The above equation may be easily solved if it includes trigonometric of the same type. If $\cos^2 x$ in the given equation is substituted by $1-\sin^2 x$, the given equation will include only $sin x$. Hence our equation becomes

    $-(1-\sin ^2 x) = -2(\sin x+1)$

  2. Expand $-1+\sin ^2 x = -2\sin x-2$

  3. Which may be written as

    $\sin ^2 x + 2\sin x+1=0$

  4. With the left side factored as follows

    $(\sin x +1)^2=0$

  5. which gives

    $\sin x +1 = 0$

    $\sin x = -1$

  6. The solution set of the above equation for $0 < x < 2\pi$ is

    $\{ \dfrac{3\pi}{2} \}$

    Answer E