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Question Number 92279 by  M±th+et+s last updated on 06/May/20

7sin(θ)+2cos^2 (θ)=5    0≤θ≤2π

$$\mathrm{7}{sin}\left(\theta\right)+\mathrm{2}{cos}^{\mathrm{2}} \left(\theta\right)=\mathrm{5} \\ $$$$ \\ $$$$\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$

Commented by jagoll last updated on 06/May/20

7sin θ+2−2sin^2 θ=5  2sin^2 θ−7sin θ+3=0  (2sin θ−1)(sin θ−3)=0  sin θ=(1/2)⇒ { ((θ=(π/6))),((θ=((5π)/6))) :}

$$\mathrm{7sin}\:\theta+\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} \theta=\mathrm{5} \\ $$$$\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{7sin}\:\theta+\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)\left(\mathrm{sin}\:\theta−\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\begin{cases}{\theta=\frac{\pi}{\mathrm{6}}}\\{\theta=\frac{\mathrm{5}\pi}{\mathrm{6}}}\end{cases} \\ $$

Commented by  M±th+et+s last updated on 06/May/20

correct solution thank you����

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