Given: [tex]2-cos2\theta=3sin\theta cos\theta[/tex] To find: [tex]tan\theta[/tex] Step-by-step explanation: Now, [tex]2-cos2\theta=3sin\theta cos\theta[/tex] [tex]\Rightarrow 2(sin^{2}\theta+cos^{2}\theta)-(cos^{2}\theta-sin^{2}\theta)=3 sin\theta cos\theta[/tex] [tex]\Rightarrow 2sin^{2}\theta+2cos^{2}\theta-cos^{2}\theta+sin^{2}\theta=3 sin\theta cos\theta[/tex] [tex]\Rightarrow 3sin^{2}\theta-3 sin\theta cos\theta+cos^{2}\theta=0[/tex] [tex]\Rightarrow 3 tan^{2}\theta-3 tan\theta+1=0[/tex] since [tex]cos\theta\neq 0[/tex] Using Sridhar Acharya’s formula, we get [tex]\quad tan\theta=\frac{3\pm\sqrt{9-12}}{6}[/tex] [tex]\Rightarrow tan\theta=\frac{3\pm\sqrt{-3}}{6}[/tex] [tex]\Rightarrow tan\theta=\frac{3\pm i\sqrt{3}}{6}[/tex] where [tex]i=\sqrt{-1}[/tex] Answer: [tex]tan\theta=\frac{3\pm i\sqrt{3}}{6}[/tex] Reply
Given:
[tex]2-cos2\theta=3sin\theta cos\theta[/tex]
To find: [tex]tan\theta[/tex]
Step-by-step explanation:
Now, [tex]2-cos2\theta=3sin\theta cos\theta[/tex]
[tex]\Rightarrow 2(sin^{2}\theta+cos^{2}\theta)-(cos^{2}\theta-sin^{2}\theta)=3 sin\theta cos\theta[/tex]
[tex]\Rightarrow 2sin^{2}\theta+2cos^{2}\theta-cos^{2}\theta+sin^{2}\theta=3 sin\theta cos\theta[/tex]
[tex]\Rightarrow 3sin^{2}\theta-3 sin\theta cos\theta+cos^{2}\theta=0[/tex]
[tex]\Rightarrow 3 tan^{2}\theta-3 tan\theta+1=0[/tex] since [tex]cos\theta\neq 0[/tex]
Using Sridhar Acharya’s formula, we get
[tex]\quad tan\theta=\frac{3\pm\sqrt{9-12}}{6}[/tex]
[tex]\Rightarrow tan\theta=\frac{3\pm\sqrt{-3}}{6}[/tex]
[tex]\Rightarrow tan\theta=\frac{3\pm i\sqrt{3}}{6}[/tex] where [tex]i=\sqrt{-1}[/tex]
Answer: [tex]tan\theta=\frac{3\pm i\sqrt{3}}{6}[/tex]