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Question Number 193381 by Mingma last updated on 12/Jun/23

Answered by som(math1967) last updated on 12/Jun/23

$$2{sin}^{2}4\theta +2{sin}^{2}2\theta -2{sin}^2\theta =1$$$$1-cos8\theta +1-cos4\theta -1+cos2\theta =1$$$$cos2\theta -(cos4\theta +cos8\theta )=0$$$$\Rightarrow cos2\theta -2cos6\theta cos2\theta =0$$$$\Rightarrow cos2\theta \{1-2cos6\theta \}=0$$$$\Rightarrow (1-2cos6\theta )=0[\theta <90]$$$$cos6\theta =\frac{1}{2}$$$$6\theta =60\therefore \theta =10$$$$$$

Commented by Mingma last updated on 12/Jun/23

Perfect ��

Commented by mnjuly1970 last updated on 13/Jun/23

$$\theta ={10}^{°},\theta ={45}^{°}$$

Commented by som(math1967) last updated on 13/Jun/23

$$yes,ifco2\theta =0\Rightarrow \theta =45$$

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