We have, 2‌cos‌θ+sin‌θ=1‌‌θ∈[−‌
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2
,‌
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2
] Squaring both sides, we get ‌⇒‌‌(2‌cos‌θ)2=(1−sin‌θ)2 ‌⇒‌‌4cos2θ=1+sin‌2θ−2sin‌θ ‌⇒‌‌5sin‌2θ−2sin‌θ−3=0 ‌⇒‌‌(sin‌θ−1)(5sin‌θ+3)=0 ‌⇒‌‌sin‌θ=1‌ and ‌−‌
3
5
‌⇒‌‌cos‌θ=0‌ and ‌+‌
4
5
‌∵‌‌7‌cos‌θ+6sin‌θ=k,‌ when ‌sin‌θ=1 ‌⇒‌‌7(0)+6(1)=k⇒k=6 When, sin‌θ=−‌