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Sagot :
We are given the following equation
[tex]\cos 2\theta-\sin \theta=0[/tex]Let us solve the above trigonometric equation.
Using the double angle identity,
[tex]\cos 2\theta=1-2\sin ^2\theta[/tex]So, the equation becomes
[tex]\begin{gathered} \cos 2\theta-\sin \theta=0 \\ 1-2\sin ^2\theta-\sin \theta=0 \end{gathered}[/tex]Now, let us solve the equation by substitution
Let sinθ = u
[tex]\begin{gathered} 1-2\sin ^2\theta-\sin \theta=0 \\ 1-2u^2-u=0 \\ -2u^2-u+1=0 \end{gathered}[/tex]Let us solve the above equation using the quadratic formula
[tex]u=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]The coefficients are
a = -2
b = -1
c = 1
[tex]\begin{gathered} u=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(-2)(1)}}{2(-2)} \\ u=\frac{1\pm\sqrt[]{1+8}}{-4} \\ u=\frac{1\pm\sqrt[]{9}}{-4} \\ u=\frac{1\pm3}{-4} \\ u=\frac{1-3}{-4},\; \; u=\frac{1+3}{-4} \\ u=\frac{-2}{-4},\; \; u=\frac{4}{-4} \\ u=\frac{1}{2},\; \; u=-1 \end{gathered}[/tex]So, the two possible values are u = 1/2 and u = -1
Substitute them back into sinθ = u
[tex]\begin{gathered} \sin \theta=\frac{1}{2},\; \; \sin \theta=-1 \\ \theta=\sin ^{-1}(\frac{1}{2}),\; \; \theta=\sin ^{-1}(-1) \\ \theta=\frac{\pi}{6}\; and\; \frac{5\pi}{6},\; \; \theta=\frac{3\pi}{2}\; \\ \theta=30\degree\; and\; \; 150\degree,\; \; \theta=270\degree \end{gathered}[/tex]Therefore, the two solutions of the given equation are θ = 30°, θ = 150°, θ = 270° on the interval [0, 360°)
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